NCERT Solutions for Class 12 Maths Chapter 13: Probability

 

It is given that, P(A|B) = P(B|A)

⇒ P (A) = P (B)

Therefore, option (D) is correct.

 

It is given that,P(A)=6/11, P(B) =5/11 and P(A ∪ B) =7/11

 

It is given that P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.2

 

Given: P (B) = 0.5, P (A ∩  B) = 0.32

 

 

 

If a coin is tossed three times, then the sample space S is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that the sample space has 8 elements.

(i) E = {HHH, HTH, THH, TTH}

F = {HHH, HHT}

∴ E ∩ F = {HHH}

(ii) E = {HHH, HHT, HTH, THH}

F = {HHT, HTH, HTT, THH, THT, TTH, TTT}

∴ E ∩ F = {HHT, HTH, THH}

(iii) E = {HHH, HHT, HTT, HTH, THH, THT, TTH}

F = {HHT, HTT, HTH, THH, THT, TTH, TTT}

 

 

If two coins are tossed once, then the sample space S is

S = {HH, HT, TH, TT}

(i) E = {HT, TH}

F = {HT, TH}

(ii) E = {HH}

F = {TT}

∴ E ∩ F = Φ

P (F) = 1 and P (E ∩ F) = 0

∴ P(E|F) =

Since a dice has six faces. Therefore E = 6 x 6 x 6 = 216

E = (1, 2, 3, 4, 5, 6) x (1, 2, 3, 4, 5, 6) x (4)

F = (6) x (5) x (1, 2, 3, 4, 5, 6)

 

 

If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be

S = {MFS, MSF, FMS, FSM, SMF, SFM}

⇒ E = {MFS, FMS, SMF, SFM}

F = {MFS, SFM}

∴ E ∩ F = {MFS, SFM}

E : Son on one end

 

Let the first observation be from the black die and second from the red die.

When two dice (one black and another red) are rolled, the sample space S has 6 × 6 = 36 number of elements.

(a) Let A: Obtaining a sum greater than 9

= {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}

B: Black die results in a 5.

= {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

∴ A ∩ B = {(5, 5), (5, 6)}

The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P (A|B).

(b) E: Sum of the observations is 8.

= {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

F: Red die resulted in a number less than 4.

The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, is given by P (E|F).

 

P (A) = 1/2  P(B) = 0

Therefore, P (A|B) is not defined.

Therefore, option (C) is correct.

 

 

When a fair die is rolled, the sample space S will be

S = {1, 2, 3, 4, 5, 6}

It is given that E = {1, 3, 5}, F = {2, 3}, and G = {2, 3, 4, 5}

(iii) E ∪ F = {1, 2, 3, 5}

(E ∪ F) ∩ G = {1, 2, 3, 5} ∩{2, 3, 4, 5} = {2, 3, 5}

E ∩ F = {3}

(E ∩ F) ∩ G = {3}∩{2, 3, 4, 5} = {3}

 

 

Let b and g represent the boy and the girl child respectively. If a family has two children, the sample space will be

S = {(b, b), (b, g), (g, b), (g, g)}

Let A be the event that both children are girls.

∴ A = {(g,g)}

(i) Let B be the event that the youngest child is a girl.

The conditional probability that both are girls, given that the youngest child is a girl, is given by P (A|B).

Therefore, the required probability is 1/2.

(ii) Let C be the event that at least one child is a girl.

The conditional probability that both are girls, given that at least one child is a girl, is given by P(A|C).

 

 

The given data can be tabulated as

Let us denote E = easy questions, M = multiple choice questions, D = difficult questions, and T = True/False questions

Total number of questions = 1400

Total number of multiple choice questions = 900

Therefore, probability of selecting an easy multiple choice question is

P (E ∩ M) = 500/1400 = 5/14

Probability of selecting a multiple choice question, P (M), is

900/1400 = 9/14

P (E|M) represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.

∴

P(E|M) = P(E ∩ M)/P(M) = 5/14/9/14 = 5/9

Therefore, the required probability is 5/9.

 

 

When dice is thrown, number of observations in the sample space = 6 × 6 = 36

Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different.
∴ A = {(1, 3), (2, 2), (3, 1)
S =  (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)

(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)

(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)

(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)

(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)

(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)

B = {(2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 2),

(3, 2), (4, 2), (5, 2), (6, 2),(1, 3), (2, 3), (4, 3),

(5, 3), (6, 3), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4),

(1, 5), (2, 5), (3, 5), (4, 5), (6, 5), (1, 6), (2, 6),

(3, 6), (4, 6), (5, 6)}

Let P (A|B) represent the probability that the sum of the numbers on the dice is 4, given that the two numbers appearing on throwing the two dice are different.

Therefore, the required probability is 1/15.

 

 

The outcomes of the given experiment can be represented by the following tree diagram.

The sample space of the experiment is,

S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

(1, H), (2, H), (3, H), (4, H), (5, H), (1, T), (2, T), (3, T), (4, T), (5, T)}

Let A be the event that the coin shows a tail and B be the event that at least one die shows 3.

Probability of the event that the coin shows a tail, given that at least one die shows 3, is given by P(A|B).

Therefore,

 

 

There are 26 black cards in a deck of 52 cards.

Let P (A) be the probability of getting a black card in the first draw.

P(A) = 26/52 = 1/2

Let P (B) be the probability of getting a black card on the second draw.

Since the card is not replaced,

P(B) = 25/51

Thus, probability of getting both the cards black = 1/2 x25/51 = 25/102

 

P (A’ and B’)

= [1 – P (A)] . [1 – P (B)]

= P (A’). P (B’)

Hence, option (B) is correct

 

It is given that P (A) = 1/2 P (B) = 7/12 and P (not A or not B) = 1/4

 

It is given that P(A) = 3/5 and P(B) = 1/5

As A and B are independent events.

P (A ∩ B) = P(A) .P(B) = 3/5.1/5 = 3/25

 

When two dice are rolled, the number of outcomes is 36.

The only even prime number is 2.

Let E be the event of getting an even prime number on each die.

∴ E = {(2, 2)}

⇒ P(E) = 1/36

Hence option (D) is correct.

 

 

Let A, B, and C be the respective events that the first, second, and third drawn orange is good.

Therefore, probability that first drawn orange is good, P (A) = 12/15

The oranges are not replaced.

Therefore, probability of getting second orange good, P (B) = 11/14

Similarly, probability of getting third orange good, P(C) = 10/13

The box is approved for sale, if all the three oranges are good.

Thus, probability of getting all the oranges good = 12/15 x 11/14 x 10/13 = 44/91

Therefore, the probability that the box is approved for sale is 44/91.

 

 

If a fair coin and an unbiased die are tossed, then the sample space S is given by,

S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

Let A: Head appears on the coin

Therefore, A and B are independent events.

 

 

When a die is thrown, the sample space (S) is

S = {1, 2, 3, 4, 5, 6}

Let A: the number is even = {2, 4, 6}

P (A) = 3/6 = 1/2

B: the number is red = {1, 2, 3}

P (B) = 3/6 = 1/2

∴ A ∩ B = {2}

Therefore, A and B are not independent.

 

Therefore, E and F are not independent events.

 

 

It is given that P(A) = 1/2 P (A ∩ B) = 3/5 and P (B) = p

(i) When A and B are mutually exclusive, A ∩ B = Φ

∴ P (A ∩ B) = 0

It is known that P(A U B) = P(A) + P(B) – P(A ∩ B)

(ii) When A and B are independent, P(A ∩ B) = P(A) .P(B) = 1/2p

It is known that, P (A U B) = P(A) + P(B) – P(A ∩ B)

 

P (A) = 0.3, P (B) = 0.4

A and B are independent events.

(i) P (A ∩ B) P (A). P (B) = 0.3 x 0.4 = 0.12

(ii) P (A U B) = P (A) + P (B) – P (A). P (B) = 0.03 + 0.4 – 0.3 x 0.4 = 0.7 – 0.12 = 0.58

(iii) P (A|B) = P (A ∩ B)/P(B) = P (A|B) = P (A) = 0.3

(iv) P (B|A) = P (A ∩ B)/P(A) = P (B|A) = P (B) = 0.4

 

It is given that, P (A) = 1/4 and P (A ∩ B) = 1/8

P(not on A and not on B) =P(A’ ∩  B’)

P (not on A and not on B) P(A U B)’ [A’ ∩ B’ = (A U B)’]

 

P (A) = 0.3, P (B) = 0.6

A and B are independent events.

(i) P (A and B) = P (A). P (B) = 0.3 x 0.6 = 0.18

(ii) P (A and not B) = P(A ∩ B’) = P (A) – P(A ∩ B) = 0.3 – 0.18 = 0.12

(iii) P (A or B) = P (A) + P (B) – P (A and B) = 0.3 + 0.6 – 0.18 = 0.9 – 0.18 = 0.72

(iv) P (neither A nor B) =  P(A’ ∩ B’) = (P(A U B’)) 1 – P (A U B) = 1 – 0.72 = 0.28

 

Probability of getting an odd number in a single throw of a die = 3/6 = 1/2

Similarly, probability of getting an even number =3/6 = 1/2

Probability of getting an even number three times = 1/2 x 1/2 x 1/2 = 1/8

Therefore, probability of getting an odd number at least once

= 1 − Probability of getting an odd number in none of the throws

= 1 − Probability of getting an even number thrice

= 1- 1/8 = 7/8

 

 

Total number of balls = 18

Number of red balls = 8

Number of black balls = 10

(i) Probability of getting a red ball in the first draw = 8/18 = 4/9

The ball is replaced after the first draw.

∴ Probability of getting a red ball in the second draw = 8/18 = 4/9

Therefore, probability of getting both the balls red = 4/9 x 4/9 = 16/81

(ii) Probability of getting first ball black =10/18 = 5/9

The ball is replaced after the first draw.

Probability of getting second ball as red = 8/18 = 4/9

Therefore, probability of getting first ball as black and second ball as red =5/9 x 4/9 = 20/81

(iii) Probability of getting first ball as red =8/18 = 4/9

The ball is replaced after the first draw.

Probability of getting second ball as black =10/18 = 5/9

Therefore, probability of getting first ball as black and second ball as red =  4/9 x 5/9 = 20/81

Therefore, probability that one of them is black and other is red

= Probability of getting first ball black and second as red + Probability of getting first ball red and second ball black

= 20/81 + 20/81 = 40/81

 

 

Probability of solving the problem by A, P (A) =1/2

Probability of solving the problem by B, P (B) =1/3

Since the problem is solved independently by A and B,

∴ P(AB) = P(A).P(B) = 1/2 x 1/3 = 1/6

P(A’) = 1 – P(A) = 1-1/2 = 1/2

P(B’) = 1 – P(B) = 1 – 1/3 = 2/3

(i)Probability that the problem is solved = P (A ∪ B)

= P (A) + P (B) − P (AB) = 1/2 + 1/3 -1/6 = 4/6 = 2/3

(ii) Probability that exactly one of them solves the problem is given by, P(A).P(B’) + P(B).P(A’)

= 1/2 x 2/3 + 1/2 x 1/3 = 1/3 + 1/6 = 1/2

 

 

(i) In a deck of 52 cards, 13 cards are spades and 4 cards are aces.

∴ P(E) = P(the card drawn is a spade) =13/52 = 1/4

∴ P(F) = P(the card drawn is an ace) = 4/52 = 1/13

In the deck of cards, only 1 card is an ace of spades.

P(EF) = P(the card drawn is spade and an ace) = 1/52

P(E) × P(F) =1/4 .1/12 = 1/52 = P(EF)

⇒ P(E) × P(F) = P(EF)

Therefore, the events E and F are independent.

(ii) In a deck of 52 cards, 26 cards are black and 4 cards are kings.

∴ P(E) = P(the card drawn is black) = 26/52 = 1/2

∴ P(F) = P(the card drawn is a king) = 4/52 = 1/13

In the pack of 52 cards, 2 cards are black as well as kings.

∴ P (EF) = P(the card drawn is a black king) =2/52 = 1/26

P(E) × P(F) = 1/2.1/13 = 1/26 =P(EF)

Therefore, the given events E and F are independent.

(iii) In a deck of 52 cards, 4 cards are kings, 4 cards are queens, and 4 cards are jacks.

∴ P(E) = P(the card drawn is a king or a queen) = 8/52 = 2/13

∴ P(F) = P(the card drawn is a queen or a jack) = 8/52 = 2/13

There are 4 cards which are king or queen and queen or jack.

∴ P(EF) = P(the card drawn is a king or a queen, or queen or a jack)

=4/52 = 1/13

P(E) × P(F) =-2/13.2/13 = 4/169 ≠ 1/13

⇒ P(E) .P(F) ≠ P(EF)

Therefore, the given events E and F are not independent.

 

Let H denote the students who read Hindi newspapers and E denote the students who read English newspapers.

It is given that,

(i) Probability that a student reads Hindi or English newspaper is

(ii) Probability that a randomly chosen student reads English newspaper, if she reads Hindi news paper, is given by P (E|H).

(iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P (H|E).

 

 

The urn contains 5 red and 5 black balls.

Let a red ball be drawn in the first attempt.

∴ P (drawing a red ball)  = 5/10 = 1/2

If two red balls are added to the urn, then the urn contains 7 red and 5 black balls.

P (drawing a red ball) = 7/12

Let a black ball be drawn in the first attempt.

∴ P (drawing a black ball in the first attempt)  = 5/10 = 1/2

If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.

P (drawing a red ball) = 5/12

Therefore, probability of drawing second ball as red is

 

 

Let E1 and E2 be the events of selecting the first bag and second bag respectively.

Let A be the event of getting a red ball.

The probability of drawing a ball from the first bag, given that it is red, is given by P (E2|A).

By using Bayes’ theorem, we obtain

 

 

Let E1 and E2 be the events that the student is a hostler and a day scholar respectively and A be the event that the chosen student gets grade A.

The probability that a randomly chosen student is a hostler, given that he has an A grade, is given by P(E1|A).

By using Bayes’ theorem, we obtain

 

 

Let E1 and E2 be the respective events that the student knows the answer and he guesses the answer.

Let A be the event that the answer is correct.

The probability that the student answered correctly, given that he knows the answer, is 1.

∴ P (A|E1) = 1

Probability that the student answered correctly, given that he guessed, is 1/4.

The probability that the student knows the answer, given that he answered it correctly, is given by P(E1|A).

By using Bayes’ theorem, we obtain

 

 

If A ⊂ B, then A ∩ B = A

⇒ P (A ∩ B) = P (A)

Also, P (A) < P (B)

Hence, option (C) is correct.

 

Let E1 = The person selected is suffering from certain disease, E2 = The person selected is not suffering from certain disease and A = The doctor diagnoses correctly

Since E1 and E2 are events complimentary to each other,

∴ P (E1) + P (E2) = 1

⇒ P (E2) = 1 − P (E1) = 1 − 0.001 = 0.999

Let A be the event that the blood test result is positive.

Probability that a person has a disease, given that his test result is positive, is given by

P (E1|A).

By using Bayes’ theorem, we obtain

 

Let E1 = a two headed coin, E2 = a biased coin, E3 = an unbiased coin and A = A head is shown

Let A be the event that the coin shows heads.

A two-headed coin will always show heads.

Probability of heads coming up, given that it is a biased coin= 75%

Since the third coin is unbiased, the probability that it shows heads is always 1/2..

The probability that the coin is two-headed, given that it shows heads, is given by

P (E1|A).

By using Bayes’ theorem, we obtain

 

Let E1 = Person chosen is a scooter driver, E2 = Person chosen is a car driver, E3 = Person chosen is a truck driver and A = Person meets with an accident

Let A be the event that the person meets with an accident.

There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers.

Total number of drivers = 2000 + 4000 + 6000 = 12000

The probability that the driver is a scooter driver, given that he met with an accident, is given by P (E1|A).

By using Bayes’ theorem, we obtain

 

Let E1 and E2 be the respective events of items produced by machines A and B. Let X be the event that the produced item was found to be defective.

∴ Probability of items produced by machine A, P (E1) = 60% = 3/5

Probability of items produced by machine B, P (E2) = 40% = 2/5

Probability that machine A produced defective items, P (X|E1)  = 2% = 2/100

Probability that machine B produced defective items, P (X|E2)  = 1% = 1/100

The probability that the randomly selected item was from machine B, given that it is defective, is given by P (E2|X).

By using Bayes’ theorem, we obtain

 

Let E1 and E2 be the respective events that the first group and the second group win the competition. Let A be the event of introducing a new product.

P (E1) = Probability that the first group wins the competition = 0.6

P (E2) = Probability that the second group wins the competition = 0.4

P (A|E1) = Probability of introducing a new product if the first group wins = 0.7

P (A|E2) = Probability of introducing a new product if the second group wins = 0.3

The probability that the new product is introduced by the second group is given by

P (E2|A).

By using Bayes’ theorem, we obtain

 

Let E1 be the event that the outcome on the die is 5 or 6 and E2 be the event that the outcome on the die is 1, 2, 3, or 4.

Let A be the event of getting exactly one head.

P (A|E1) = Probability of getting exactly one head by tossing the coin three times if she gets 5 or 6 = 3/8

P (A|E2) = Probability of getting exactly one head in a single throw of coin if she gets 1, 2, 3, or 4 = 1/2

The probability that the girl threw 1, 2, 3, or 4 with the die, if she obtained exactly one head, is given by P (E2|A).

By using Bayes’ theorem, we obtain

 

Let E1 = the item is manufactured by the operator A, E2 = the item is manufactured by the operator B, E3 = the item is manufactured by the operator C and A = the item is defective

Let X be the event of producing defective items.

The probability that the defective item was produced by A is given by P (E1|A).

By using Bayes’ theorem, we obtain

 

Let E1 = the missing card is a diamond, E2 = the missing card is a spade, E3 = the missing card is a club, E4 = the missing card is a heart and A = drawing of two heart cards from the remaining cards.

Let E1 and E2 be the respective events of choosing a diamond card and a card which is not diamond.

Let A denote the lost card.

Out of 52 cards, 13 cards are diamond and 39 cards are not diamond.

When one diamond card is lost, there are 12 diamond cards out of 51 cards.

Two cards can be drawn out of 12 diamond cards in ways.

Similarly, 2 diamond cards can be drawn out of 51 cards in ways. The probability of getting two cards, when one diamond card is lost, is given by P (A|E1).

When the lost card is not a diamond, there are 13 diamond cards out of 51 cards.

Two cards can be drawn out of 13 diamond cards in ways whereas 2 cards can be drawn out of 51 cards in ways.

The probability of getting two cards, when one card is lost which is not diamond, is given by P (A|E2).

The probability that the lost card is diamond is given by P (E1|A).

By using Bayes’ theorem, we obtain

= 11/2/25 = 11/50

 

Let A be the event that the man reports that head occurs in tossing a coin and let E1 be the event that head occurs and E2 be the event head does not occur.

E1: A speaks truth

E2: A speaks false

Let X be the event that a head appears.

P(E1)=45

Therefore,

P(E2)=1-P(E1)=1-45=15

If a coin is tossed, then it may result in either head (H) or tail (T).

The probability of getting a head is 1/2 whether A speaks truth or not.

The probability that there is actually a head is given by P (E1 | X)

Hence, option (A) is correct.

 

 

It is known that the sum of all the probabilities in a probability distribution is one.

(i) Sum of the probabilities = 0.4 + 0.4 + 0.2 = 1

Therefore, the given table is a probability distribution of random variables.

(ii) It can be seen that for X = 3, P (X) = −0.1

It is known that the probability of any observation is not negative. Therefore, the given table is not a probability distribution of random variables.

(iii) Sum of the probabilities = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1

Therefore, the given table is not a probability distribution of random variables.

(iv) Sum of the probabilities = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 ≠ 1

Therefore, the given table is not a probability distribution of random variables.

 

There two balls may be selected as BR, RB, BR, BB, where R represents red ball and B represents black ball.

Variable X has the value 0, 1, 2, i.e., there may be no black ball, may be one black ball or both the balls are black.

 

A coin is tossed six times and X represents the difference between the number of heads and the number of tails.

∴ X (6 H, 0T) = |6 – 0| = 6

X (5 H, 1 T)  = |5 – 1| = 4

X (4 H, 2 T)  = |4 – 2| = 2

X (3 H, 3 T)  = 3 – 3| = 0

X (2 H, 4 T)  = |2 – 4| = 2

X (1 H, 5 T)  = |1 – 5| = 4

X (0H, 6 T) = | 0 – 6| = 6

Thus, the possible values of X are 6, 4, 2, and 0.

 

(i) When one coin is tossed twice, the sample space is

{HH, HT, TH, TT}

Let X represent the number of heads.

∴ X (HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0

Therefore, X can take the value of 0, 1, or 2.

It is known that,

P(HH) = P(HT) = P(TH) = P(TT) = 1/4

P (X = 0) = P (TT) = 1/4

P (X = 1) = P (HT) + P (TH) = 14 + 1/4 = 1/2

P (X = 2) = P (HH) = 1/4

Thus, the required probability distribution is as follows.

(ii) When three coins are tossed simultaneously, the sample space is {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}

Let X represent the number of tails.

It can be seen that X can take the value of 0, 1, 2, or 3.

P (X = 0) = P (HHH) = 1/8

P (X = 1) = P (HHT) + P (HTH) + P (THH) = 1/8 + 1/8 + 1/8 = 3/8

P (X = 2) = P (HTT) + P (THT) + P (TTH) =  1/8 + 1/8 + 1/8 = 3/8

P (X = 3) = P (TTT) = 1/8

Thus, the probability distribution is as follows.

(iii) When a coin is tossed four times, the sample space is

(iii) When a coin is tossed four times, the sample space is

S = {HHHH,HHHT,HHTH,HHTT,HTHT,HTHH,HTTH,HTTT,THHH,THHT,THTH,THTT,TTHH,TTHT,TTTH,TTTT}

Let X be the random variable, which represents the number of heads.

It can be seen that X can take the value of 0, 1, 2, 3, or 4.

P (X = 0) = P (TTTT) = 1/16

P (X = 1) = P (TTTH) + P (TTHT) + P (THTT) + P (HTTT)

= 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4

P (X = 2) = P (HHTT) + P (THHT) + P (TTHH) + P (HTTH) + P (HTHT)

+ P (THTH)

= 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8

P (X = 3) = P (HHHT) + P (HHTH) + P (HTHH) P (THHH)

= 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4

P (X = 4) = P (HHHH) = 1/16

Thus, the probability distribution is as follows.

 

When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.

Let X be the random variable, which represents the number of successes.

(i) Here, success refers to the number greater than 4.

P (X = 0) = P (number less than or equal to 4 on both the tosses) = 4/6 X 4/6 = 4/9

P (X = 1) = P (number less than or equal to 4 on first toss and greater than 4 on second toss) + P (number greater than 4 on first toss and less than or equal to 4 on second toss)

= 4/6 X 2/6 + 4/6 X 2/6 = 4/9

P (X = 2) = P (number greater than 4 on both the tosses)

= 2/6 X 2/6 = 1/9

Thus, the probability distribution is as follows.

(ii) Here, success means six appears on at least one die.

P (Y = 0 ) = P (six appears on  none of the dice) = 5/6 x  5/6 = 25/36

P (Y = 1) = P (six appears on at least one of the dice) = 1/6 x 5/6 + 5/6 x 1/6 + 1/6 x 1/6 = 11/36

Thus, the required probability distribution is as follows.

 

It is given that out of 30 bulbs, 6 are defective.

⇒ Number of non-defective bulbs = 30 − 6 = 24

4 bulbs are drawn from the lot with replacement.

Let X be the random variable that denotes the number of defective bulbs in the selected bulbs.

Therefore, the required probability distribution is as follows.

 

 

Let the probability of getting a tail in the biased coin be x.

∴ P (T) = x

⇒ P (H) = 3x

For a biased coin, P (T) + P (H) = 1

When the coin is tossed twice, the sample space is {HH, TT, HT, TH}.

Let X be the random variable representing the number of tails.

∴ P (X = 0) = P (no tail) = P (H) × P (H) = 3/4 x 3/4 = 9/16

P (X = 1) = P (one tail) = P (HT) + P (TH)

= 3/4 . 1/4 + 1/4 . 3/4

= 3/16 + 3/16= 3/8

P (X = 2) = P (two tails) = P (TT) = 1/4 x 1/4 = 1/16

Therefore, the required probability distribution is as follows.

 

(i) Since, the sum of all the probabilities of a distribution is 1.

 

(a) It is known that the sum of probabilities of a probability distribution of random variables is one.

∴ k + 2k + 3k + 0 = 1

⇒ 6k = 1

⇒ k =1/6

(b) P(X < 2) = P(X = 0) + P(X = 1)

= k + 2k

= 3k

=3/6 = 1/2

 

Let X denote the success of getting heads.

Therefore, the sample space is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that X can take the value of 0, 1, 2, or 3

Therefore, the required probability distribution is as follows.

 

Two dice thrown simultaneously is the same as the die thrown 2 times.

Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1, or 2.

∴ P (X = 0) = P (not getting six on any of the dice) = 25/36

P (X = 1) = P (six on first die and no six on second die) + P (no six on first die and six on second die)

=2(1/6 x 5/6) = 10/36

P (X = 2) = P (six on both the dice) =1/36

Therefore, the required probability distribution is as follows.

Then, expectation of X = E(X) =∑ X iP(Xi)

= 0 x 25/36 + 1 x 10/36 + 2 x 1/36

= 1/3

 

The two positive integers can be selected from the first six positive integers without replacement in 6 × 5 = 30 ways

X represents the larger of the two numbers obtained. Therefore, X can take the value of 2, 3, 4, 5, or 6.

For X = 2, the possible observations are (1, 2) and (2, 1).

P(X = 2) = 2/30 = 1/15

For X = 3, the possible observations are (1, 3), (2, 3), (3, 1), and (3, 2).

P(X = 3) = 4/30 = 2/15

For X = 4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2), and (4, 1).

P(X = 4) = 6/30 = 1/5

For X = 5, the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2), and (5, 1).

P(X = 5) = 8/30 = 4/15

For X = 6, the possible observations are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6,5) , (6, 4), (6, 3), (6, 2), and (6, 1).

P(X = 6) = 10/30 = 1/3

Therefore, the required probability distribution is as follows.

 

When two fair dice are rolled, 6 × 6 = 36 observations are obtained.

P(X = 2) = P(1, 1) = 1/36

P(X = 3) = P (1, 2) + P(2, 1) = 2/36 = 1/18

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 3/36 = 1/12

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) = 4/36 = 1/9

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = 5/36

P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = 6/36 = 1/6

P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = 5/36

P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) = 4/36 = 1/9

P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4) = 3/36 = 1/12

P(X = 11) = P(5, 6) + P(6, 5) = 2/36 = 1/18

P(X = 12) = P(6, 6) = 1 x 36

Therefore, the required probability distribution is as follows.

 

There are 15 students in the class. Each student has the same chance to be chosen. Therefore, the probability of each student to be selected is 1/15.

The given information can be compiled in the frequency table as follows.

P(X = 14) =2/15, P(X = 15) =1/15, P(X = 16) =2/15, P(X = 16) =3/15,

P(X = 18) =115, P(X = 19) =2/15, P(X = 20) =3/15, P(X = 21) =1/15

Therefore, the probability distribution of random variable X is as follows.

Then, mean of X = E(X)

 

 

It is given that P(X = 0) = 30% = 30/100 = 0.3

P(X = 1) = 70% = 70/100 = 0.7

Therefore, the probability distribution is as follows.

= 0.7 − (0.7)2

= 0.7 − 0.49

= 0.21

 

Let X be the random variable representing a number on the die.

The total number of observations is six.

Therefore, the probability distribution is as follows.

Therefore, option (B) is correct.

 

Let X denote the number of aces obtained. Therefore, X can take any of the values of 0, 1, or 2.

In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.

Therefore, option (D) is correct.

 

Let X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials.

It is given that, p = 0.05

 

Let X represent the number of spade cards among the five cards drawn. Since the drawing of cards is with replacement, the trials are Bernoulli trials.

In a well shuffled deck of 52 cards, there are 13 spade cards.

 

The repeated tosses of a die are Bernoulli trials. Let X denote the number of successes of getting odd numbers in an experiment of 6 trials.

Probability of getting an odd number in a single throw of a die is, p – 3/6 = 1/2

q = 1 – p = 1/2

X has a binomial distribution.

 

The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.

Probability of getting doublets in a single throw of the pair of dice is

 

Let X denote the number of defective items in a sample of 10 items drawn successively. Since the drawing is done with replacement, the trials are Bernoulli trials.

 

Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.

Since the balls are drawn with replacement, the trials are Bernoulli trials.

X has a binomial distribution with n = 4 and p = 1/10

 

Let X represent the number of correctly answered questions out of 20 questions.

The repeated tosses of a coin are Bernoulli trails. Since “head” on a coin represents the true answer and “tail” represents the false answer, the correctly answered questions are Bernoulli trials.

 

X is the random variable whose binomial distribution is B(6,1/2).

Therefore, n = 6 and p = 1/2

Therefore, P (X = 3) is maximum.

 

The repeated guessing of correct answers from multiple choice questions are Bernoulli trials. Let X represent the number of correct answers by guessing in the set of 5 multiple choice questions.

Probability of getting a correct answer is, p = 1/3

 

Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli trials.

Clearly, X has a binomial distribution with n = 50 and p = 1/100

 

The repeated tossing of a die are Bernoulli trials. Let X represent the number of times of getting 5 in 7 throws of the die.

Probability of getting 5 in a single throw of the die, p = 1/6

 

The repeated tossing of the die are Bernoulli trials. Let X represent the number of times of getting sixes in 6 throws of the die.

Probability of getting six in a single throw of die, p = 1/6

Let A represents the favourable event i.e., 6

 

The repeated selections of articles in a random sample space are Bernoulli trails. Let X denote the number of times of selecting defective articles in a random sample space of 12 articles.

Clearly, X has a binomial distribution with n = 12 and p = 10% = 10/100 = 1/10

 

 

option (C) is correct.

The repeated selection of students who are swimmers are Bernoulli trials. Let X denote the number of students, out of 5 students, who are swimmers.

Probability of students who are not swimmers, q = 1/5

 

Therefore, option (A) is correct.