Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability are part of Plus Two Maths Chapter Wise Questions and Answers. Here we have given Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability.

Board | SCERT, Kerala |

Text Book | NCERT Based |

Class | Plus Two |

Subject | Maths Chapter Wise Questions |

Chapter | Chapter 13 |

Chapter Name | Probability |

Number of Questions Solved | 49 |

Category | Kerala Plus Two |

## Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

**Short Answer Type Questions (Score 3)**

Question 1.

a. A coin is tossed 6 times. What is I the probability of getting exactly 4 heads. **(2)**

b. Find the probability of obtaining an even prime number as the sum when a pair of dice are rolled.** (1)**

Answer:

a. In binomial distribution with n = 6

b. Let E be the event of getting a even prime number as sum is E = (1, 1)

total outcome = 36

Question 2.

In an examination, 30% of students failed in Maths, 20% of students failed in Chemistry and 10% of students failed in both Maths & Chemistry. A student is selected at random. Find the probability that:

1. The student has failed in Maths. **(1)**

2. The student has failed in Chemistry. **(1)**

3. The student has failed in Maths, it is known that he has failed in Chemistry. **(1)**

Answer:

Question 3.

A die is thrown and six possible outcomes are equally likely. E is the event, the number appearing is a multiple of 3, and F is the event; ‘The number appearing is even’. Calculate

(i) P(E) and P(F) **(1)**

(ii) P(E ∩ F) **(1)**

(iii) Prove that the events E and F are independent.**(1)**

Answer:

Question 4.

Find the probability distribution of the number of successes in two tosses of a die where a success is defined as a number greater than 4. **(3)**

Answer:

Let ‘X’ denote a random variable which is the number of successes (getting a number > 4) obtained in two tosses of a die. Clearly X takes the values 0,1, or 2.

Question 5.

Find the bionomial distribution whose mean is 10 and standard deviation **(3)**

Answer:

Let (q + p)^{n} be the required Binomial Distribution. Given np = mean of the distribution =10…(1)

Question 6.

A bag ‘X’ contains 2 white and 3 red balls and a bag Y contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag Y. **(3)**

Answer:

Let E_{1} the bag X is chosen

E_{2 }the bag Y is chosen A : the ball is red

P(E_{2}/A) is to be obtained

Question 7.

a. 5 boys and 5 girls are sitting in a row randomly. Find the probability that boys and girls sit alternatively.

b. If P(A) = , P(B) = and

P(A∪B) = then the events A and B are

( a. Dependent only on A

b. Mutually exclusive

c. Independent

d. Dependent only on A ) **(2)**

Answer:

a.

There are two cases that (1) Left end start with boy and (2) Left end start with a girl.

in 1^{st} case boys can arranged = B = 5! x 5!

in 2^{nd} case girls can arranged = G = 5! x 5!

total way 2 x 5! x 5!

sample space is 10!

∴ P = (B+G/BG) =

b.

Independent P (A) = , P (B) =

P(A∪B)=

P (A) + P (B)

P (A∪B) = P (A) + P (B)

here P (A∩B) = 0

∴ A & B are independent events

Question 8.

a. What is the probability that the position in which the consonants remains unchanged when the letters of “MATH” are re-arranged? **(1)**

b. A shooter can hit a target once in a 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target. **(2)**

Answer:

a. The total number of ways the word ‘MATH’ can be arranged = 4!

The consonants M, T, H are re-arranged as 3! ways

∴ required probability =

b. Probability of hitting the target in one shot

Probability of missing the target in one shot =

probability of hitting at least one =

=

Question 9.

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that one of them is black and other is red. **(3)**

Answer:

Let A be the event of getting a red ball. We require P (AB) + P (BA)

= P (A) P (B) + P (B) P (A)

Question 10.

Out of 9 outstanding students of a school, there are 4 boys and 5 girls. A team of 4 students is to be selected for a quiz competition. Find the probability that 2 boys and 2 girls are selected. **(3)**

Answer:

Out of 4 students 4 can be selected in , ways favourable cases of selecting 2 boys and 2 girls out of 4 boys and 5

Probability of selecting 2 boys and 2 girls out of 9

**Long Answer Type Questions**

**(Score 4)**

Question 1.

In a binomial distribution mean is 6 and standard deviation is What is the number of trials. **(4)**

Answer:

Mean np = 6

Question 2.

A company has two plants to manufacture scooters. Plant I manufactures 70% of the scooters and Plant II manufactures 30%. At Plant I, 80% of the scooters are of standard quality and at Plant II, 90% of the scooters are rated as of standard quality. A scooter is chosen at random and is found to be of standard quality. What is the probability that it has come from Plant II? **(4)**

Answer:

Let E_{1} – Plant I is chosen

E_{2} – Plant II is chosen

A – Scooter is of standard quality

Question 3.

Let E and F be events such that

Answer:

Question 4.

A bag contains 4 white and 2 black balls. Another bag contains 3 white and 5 black balls. If one ball is drawn from each bag. find the probability that

i. Both are white. **(2)**

ii. Both are black. **(1)**

iii. One is white and one is black. **(1)**

Answer:

W_{1} = Drawing a white ball from I -bag

W_{2} = Drawing a white ball from II -bag

B_{1 } = Drawing a black ball from I -bag

B_{2} = Drawing a black ball from II -bag

i. P(Both are white) = P(W1,W_{2})

= P(W_{1}).P(W_{2}) =

ii. P(Both black) = P(W_{1}).P(W_{2}) =

iii. P[one white & one black] = P [Black from I and white from second] + P[white from I and black from II]

= P(W_{1}).P(W_{2}) =

Question 5.

It is known that 10% of certain articles manufactured are defective. What is I the probablity that in a random sample of 12 such articles, 9 are defective? **(4)**

Answer:

This is a Bernoulli trial with n= 12

Here success is selecting a defective item.

P = probability of success

Question 6.

A can hit target 4 times out of 5 times, B can hit target 3 times out of 4 times and C can hit target 2 times out of 3 times. They fire simultaneously. Find the probability.

i. Any two out of A,B and C will hit the target. **(2)**

ii. None of them will hit the target. **(2)**

Answer:

Question 7.

A Coin is tossed 5 times.

i. What is the probability Of getting 3 heads? **(2)**

ii. What is the probability of getting at least 3 heads? **(2)**

Answer:

Let P denote the probability of getting head in a single toss of a coin.Then

P=;q =

Let ‘X’ denote the number of heads in 5 tosses of a coin (n = 5).

Question 8.

Assume that each child born is equally likely to be boy or a girl. If a family has two children, what is the conditional probability that both are girls given that the youngest is a girl ? **(4)**

Answer:

Let b and g represent the boy and the girl child, respectively. If a family has two children, the sample space will be S= {bb, bg,gb,gg}

which contains four equally likely sample points ie.,n (s) = 4

Question 9.

A coin is biased such that a head is three times as likely to occur than a tail. When it is tossed twice, find the probability distribution of number of heads. **(4)**

Answer:

Let probability of getting a tail = p

∴ Probability of getting a head = 3p

[given that, head is three times as likely to occur than a tail]

3p+p=1

[total probability of all outcomes of an experiment is 1]

4p = 1 ⇒ p=

P(getting a tail) = p =

and P(getting a head) = 3p =

Now, let X = number of heads when the coin is tossed twice.

So, X can take values 0,1 and 2.

Now, P(X=0) = P(no head occurs)

Now; P(X=1) = P(one head occurs)

Question 10.

Assume that each child born is equally likely to be boy or a girl. If a family has two children, what is the conditional probability that both are girls given that atleast one is a girl? **(4)**

Answer:

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

S= {bb, bg,gb,gg}

which contains four equally likely sample points ie.,n (s) = 4

Let E: both children are girls, then

**Very Long Answer Type Questions (Score 6)**

Question 1.

a. A and B independently try to solve a problem. Probability that A solves the problem is 1/3 and that B solves the problem is 1/4. Find the probability that

i. Both of them solves the problem. **(1)**

ii. The problem is solved. **(2)**

b. Find the probability distribution of the number of heads in three tosses of a fair coin.**(3)**

Answer:

Question 2.

Find the probability distribution of the number of heads in three tosses of a fair coin.

Answer:

a. Two events A and B are said to be mutually exclusive events if

P(A∩B) = 0 P(A∩B)=θ **(2)**

b. Let A = Event of drawing a one rupee coin . **(4)**

B_{1} = Event of selecting I-compartment

B_{2} =Event of selecting II-compartment

P(A∩B_{1}) = P(B_{1}).P(A/B_{1}) =

P(A∩B_{2}) = P(B_{2}).P(A/B2) =

Required probability

Question 3.

A random variable X has the following probability distribution. **(6)**

X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

P(X) | 0 | k | 2k | 2k | 3k | k^{2} | 2k^{2} | 7k^{2}+k |

Determine

i. k

ii. P (X<3)

iii. P (X>6)

iv. P (0<X <3)

Ans.

i.

X is a random variable P (X=X_{1}) ≥ 0,

P (X) > 0 and ΣP (X)=l

0+k+2k+2k+3 k +k^{2}+2 k ^{2}+7 k ^{2}+ k =1

10k^{2} +9 k -1=0 .

(k+1) (10k-1) = 0

k=-1 or 1 =

Since k is not negative, K=

ii. P (X<3)

= P (X =0) + P (X=l) + P (X=2)

= 0+k+2k+3k=

iii. P (X>6) = P (X=7) = Ik ^{2}+k

=

iv. P (0< X<3) = P (X=1)+ P (x=2)

= k +2 k =3 ,k =

Question 4.

In four throws with a pair of dice, what is the probability of throwing doublets.

(i) Twice **(3)**

(ii) At least twice. **(3)**

Answer:

A pair of dice can be thrown in 6 x 6 = 36 ways.

Doublets can be obtained as

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

Question 5.

Suppose that 5 men out of 100 and 25 women out of thousand are colour blind. A colour blind person is chosen at random. What is the probability of this being male (assuming that males and females are in equal proportion).**(6)**

Answer:

Let us define a few events

M: A person is male

F: A person is female

C: A person is colour blind.

Obviously, we want to calculate P (M/C).

Also it is given that P (M) ,P(F) = each,

as males and females are in equal proportion,

Question 6.

A can hit a target 4 times in 5 shots. B three times in 4 shots, and C two times in three shots. Calculate the probability that

i. A, B, C all may hit. **(2)**

ii. B, C may hit and A may lose. **(1)**

iii. Any two will hit the target. **(2)**

iv. None of them will hit the target.**(1)**

Answer:

Question 7.

An urn contains 5 white and 8 black balls. Two successive draws of three balls at a time are made such that

i. The balls are replaced before the second trial.

ii. The balls are not replaced before the second trial.

iii. Find the probability in each case that the first drawing will give 3 white and the second 3 black balls. **(6)**

Answer:

Let E = Event of drawing three white balls in I – draw.

F = Event of drawing three black balls in II – draw.

(i) The balls are replaced

Question 8.

Two persons A and B throw a die alternatively till one of them gets a three and wins the game. If A begins, find

i. the probability of winnings of A. **(3)**

ii. the probability of winnings of B. **(3)**

Answer:

(i) Let E = A gets a three

F = B gets a three

A wins if he throws a three in the 1^{st }or 3^{rd} or 5^{th} ….. throws.

P(three in first row) =1/6

A will get a third throw if he fails in the first and B fails in second.

P(A wins in third throw)

Question 9.

A doctor claims that 60% of the patients he examines are allergic to some type of weed, what is the probability that

i. Exactly three of his next four patients are allergic to weeds. **(3)**

ii. None of his next four patient is allergic to weeds. **(3)**

Answer:

Let A = First patient is allergic

B = Second patient is allergic

C = Third patient is allergic

D = Fourth patient is allergic

P(A)=P(B**)=**P(C)=P(D)= =

Question 10.

Probabilities that a problem will be solved by three students are respectively. What is the probability that

i. All solves the problem.

ii. None solves the problem in.

iii. The problem will be solved

iv. Exactly one solves the problem **(6)**

Answer:

**Edumate Questions & Answers**

Question 1.

A coin is tossed three times, where the events

A: occurring at most two heads ,

B: occurring at most one tail

i. Write P(A), P(B).

ii. Find P(A|B) and P(B|A)

Answer:

i. S = { TTT , JTH, THT, HU,

THH, HHT, HTH, HHH }

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

B={HHH, HHT, THH, HTH)

Question 2.

In a hostel 50% of the girls like tea , 40% and coffee. A girl is selected at random.

i. Find the probability that she likes neither tea nor coffee.

ii. If the girl likes tea , then find the probability that she likes coffee.

iii. If she likes coffee then find the probability she likes tea.

Answer:

Let T denotes the set of girls who like tea and C denotes who like coffee.

Question 3.

Vineetha and Reshma are competing for the post of school leader . The probability Vineetha to be elected is 0.6 and Reshma is 0.4. Further if Vineetha is elected the probabilityof introducing a new pattern of election 0.7 and the corresponding probability is 0.3, if Reshma is elected. Find the probability that the new pattern of election is introduced by is introduced by Reshma.

Answer:

Let E_{1 }and E_{2} be the respectively probability that Vineetha and Reshma will be elected. Let Abe the probability that a new pattern of election is introduced.

Question 4.

Find the probability distribution of the number of success in two tosses of a die where the success is defined as getting a number less than 5.

Answer:

Here success refers to the number less than 5.

Therefore the Random variable can take the values 0,1,2

P(X = 0) = P (getting the number greater than or equal to 5 on both tosses)

P(X = 1) = P (getting the number greater than or equal to 5 on first toss or getting the number less than 5 on second toss)

= (getting the number less than 5 on second toss)

=

Question 5.

i. Given that E and F are events such that P(E)=0.6, P(F) = 0.4 and p(E∩F)=0.2

ii. A die thrown and the sum of the numbers appearing is observed to be greater than 9. What is the conditional probability that the number 5 is appeared at least once.

Answer:

Question 6.

i. A and B are two events such that and .Find P(B) if they are mutually exclusive .

ii. A box contains 3 red and 4 blue balls. Two balls are drawn one by one without replacement .Find the probability of getting both balls red.

Answer:

i.

ii. Let A be the event that the first ball drawn is red and B be the event of drawing red ball in the second draw

P(A)

= Probability of getting one red ball in the second draw

P(A∩B)=P(A).

ii. Let Q denote the event that the card drawn is Queen K denote the event of drawing a king

iii. Let Q denote the event that the card drawn is Queen K denote the event of drawing a king

Question 7.

i. Find P(A∩B), If A and B are independent events with P(A) =

ii. An unbiased die is throw twice .let event of an even number in the second throw .Check the

independence of the events A and B.The probability of solving a problem independently by A and B are and respectively . Find the probability that exactly one them solves the problem.

Answer:

i.

Question 8.

i. A set of events E_{1},E_{2},……. E_{n} are said to be a partition of the sample space,then which of the following conditions is always not true

ii. A person has undertaken a business. The probabilities are 0.80 that there will be a crisis,0.85 that the business will be completed on time if there is no crisis and 0.35 that the business will be completed on time if there is a crisis. Determine the probability that the business will be completed on time.

iii. A box contains 5 red and 10 black balls. A ball is drawn at random , its colour is noted and is returned to the box. More over 2 additional balls of the colour drawn are put in the box and then a ball is drawn. What is the probability that the second ball is red?

Answer:

i. P(E_{1})≥P(E_{n})

ii. Let A be the event that the business will be completed on time and B be the event that there will be a crisis.

P(B) = 0.80

P(no crisis) = P(B^{1}) = 1 – P(B) = 0.20

P(A/B) = 0.35

P(A / B) = 0.85

By theorem on total probability

= 0.8x 0.35 + 0.20 x 0.85 = 0.45

iii. Let a red ball be drawn in the first attempt

P(drawing a red ball) =

If two red balls are added to thew box, then the box contains 7 red balls and 10 black halls.

P(darwing a red ball) =

Let a black ball be drawn in the first attempt

P(drawing a black ball) =

If two black balls are added to the box, then the box contains 5 red and 12 black balls

P(drawing a red ball) =

Probability of drawing the second ball red is =

Question 9.

i. Bag I contains 5 red and 6 black balls. Bag II contains 7 red and 5 black ; balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag I.

ii. A card from a pack of 52 cards is lost, From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being diamond.

Answer:

i. Let E_{1} be the event of choosing Bag I

and B_{2} be the event of choosing Bag II.

A be the event of drawing a red ball

ii. Let E, be the event of choosing a diamond and E_{2} be the event of choosing a non diamond card A be the event that a card is lost

When a diamond card is lost, there are 12 diamond cards in 52 cards. Then

When a non diamond card is lost, there are 13 diamond in 51 cards. Then

Question 10.

i. If X denotes number of heads obtained in tossing two coins. Then which of the following is false

(X (HH) = 2, X(HT)=1,

X(TH) = 0, X(TT) = 0)

ii. Find the probability distribution of the number of heads in the simultaneous toss of two coins.

iii. A coin is tossed so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Answer:

i. X(TH) = 0

ii. Sample space is S = {HH, HT, TH, TT}

Let X denote the number of heads, then

X(HH) = 2, X(HT) = 1, X(TH) = 1

X (TT) = 0

Therefore X can take the values 0,1 or 2.

Question 11.

i. Fill in the blank of a probability distribution of a random variable X.

X 0 1 2

P(X) 0.2 0.4 ……….

(0.2,0 .3, 0.4, 0.6)

ii. Find the mean and variance of the number of headsin three tosses of a coin.

Answer:

i. 0.4 .

ii. Let X denote the number of heads in three tosses of a coin. Sample space is

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

Then X can take the values 0, 1,2 or 3.

P(X=0)=P(TTT) = P(T)P(T)P(T)

P(X=1) = P(HTT)+P(THT)+P(TTH)

Question 12.

i. The probability that a student is not a sportsman is 1/5. Then the probability that out of 6 students, 4 are sportsman is …

ii. Find the probability of getting almost 2 sixes in 6 throws of a single die.

Answer:

ii.

The repeated tossing of the die are Bemoullis trails. Let X represents the number of times of getting sixes in 6 throws of the die.

**NCERT Questions & Answers**

Question 1.

On a multiple choice examination with three possible answers (out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

Answer:

Let p denote the probability of only one correct answer out of the three possible answers.

Let X denote the number of correct answer for each of the five questions. Then,

Probability of getting four or more correct answers

= P(X>4) =p(X=4)+P(X=5)

Question 2.

A and B are two events associated with a random experiment such that P(A) = 0.8, P (B) = 0.5, P(B|A) = 0.4. Find

i. P(A ∩B)

ii. P(A|B)

iii. P(A∪B)

Answer:

Question 3.

Two cards are drawn without replacement from a well shuffled pack of 52 cards. Find the probability that one is a spade and other is a queen of red colour.

Answer:

S_{i} =getting spade card in the i^{th} draw (i= 1,2)

Q_{i}=getting red queen card in the i^{th} draw (7=1,2).

Required Probability

Question 4.

Find the probability distribution of ‘X’, the number of heads in two tosses of a coin (a simultaneous toss of two coins).

Answer:

They may be 1 head,2 heads or no head at all. Thus the possible values of X are 0,1,2.

Question 5.

An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be I atleast 4 successes.

Answer:

Let p be the probability of a success and i q the probability of failure, Then p + q = 1 and p = 2q

Solving p = and q=

Let X be the number of success.

Then X is a binomial distribution with

Question 6.

A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.

Answer:

This is a Bernoulli trial with n = n

Here success is getting 6

P = probability of success =

Question 7.

An electronic assembly consists of two sub systems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known.

P (A fails) = 0.2

P (B fails alone) = 0.15

P (A and B fail) = 0.15

Evaluate the following probabilities

i. P (A fails | B has failed)

ii. P (A fails alone)

Answer:

P(B fails) = P (B fails alone) + P (both A & B fail) = 0.15 + 0.15 = 0.30

P (A fails |B has failed)

ii. P(A fails) = P(A fails alone) + P(both A and B fail)

0.2 = P (A fails alone) + 0.15

∴ P(A fails alone) = 0.2 – 0.15 = 0.05

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