12th NCERT Probability Exercise 13.4 Questions 17

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Question (1)

State which of the following are not the probability distributions of a random variable. Give reasons for your answer. (i)x | 0 | 1 | 2 |

p(x) | 0.4 | 0.4 | 0.2 |

Solution

∑ p(x) = 0.4 + 0.4 + 0.2= 1.

Yes it is probability distribution function.

(ii)

x | 0 | 1 | 2 | 3 | 4 |

P(x) | 0.1 | 0.5 | 0.2 | -0.1 | 0.3 |

Solution

As one of the probability is negative , it is not probability distribution function.(iii)

x | -1 | 0 | 1 |

p(x) | 0.6 | 0.1 | 0.2 |

Solution

∑p(x) = 0.6 + 0.1+ 0.2 = 0.9 < 1As the sum is less than 1, so no it is not probability distribution function.

(iv)

z | 3 | 2 | 1 | 0 | -1 |

p(x) | 0.3 | 0.2 | 0.4 | 0.1 | 0.05 |

Solution

∑ p(x) = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 > 1As sum of probabilities is greater than 1, so it is not probability distribution function.

Question (2)

An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?Solution

in an urn 5 red + 2 black = 7 ballsTwo balls drawn at random. n = 7C

X ; the number of black balls

X = 0, 1, 2.

\[P\left( {x = 0} \right) = \frac{{5{C_2}}}{{7{C_2}}} = \frac{{5 \times 4}}{{7 \times 6}} = \frac{{10}}{{21}}\] \[P\left( {x = 1} \right) = \frac{{5{C_1} \times 2{C_1}}}{{21}} = \frac{{5 \times 2}}{{21}} = \frac{{10}}{{21}}\] \[P\left( {x = 2} \right) = \frac{{2{C_2}}}{{21}} = \frac{1}{{21}}\] \[\sum {p(x) = \frac{{10}}{{21}} + \frac{{10}}{{21}} + \frac{1}{{21}} = 1} \] So yes it is probability distribution function.

Question (3)

Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?Solution

X ; difference between number of heads and number of tails when coin is tossed 6 times.No. of heads | No. of tails | Difference. |

0 | 6 | 6 |

1 | 5 | 4 |

2 | 4 | 2 |

3 | 3 | 0 |

4 | 2 | 2 |

5 | 1 | 4 |

6 | 0 | 6 |

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

Question (4)

Find the probability distribution of (i) number of heads in two tosses of a coin (ii) number of tails in the simultaneous tosses of three coins (iii) number of heads in four tosses of a coinSolution

(i) The coin is tossed two times.So sample space S = { HH, HT, TH, TT}. n = 4

X ; number of heads in two tosses of a coin.

X = 0, 1, 2.

P(X = 0) = 1/4., P(X = 1) = 2/4, P(X = 2) = 1/4.

The probability distribution function is as follows.

X | 0 | 1 | 2 | Total |

p(X) | 1/4 | 2/4 | 1/4 | 1 |

(ii) The coin is tossed three times.

So sample space S = { HHH, HHT,HTH, HTT, THH,THT, TTH TTT}. n = 8

X ; number of heads in three tosses of a coin.

X = 0, 1, 2, 3.

P(X = 0) = 1/8, P(X = 1) = 3/8, P(X = 2) = 3/8, P(X = 3) = 1/8.

The probability distribution function is as follows.

X | 0 | 1 | 2 | 3 | Total |

p(X) | 1/8 | 3/8 | 3/8 | 1/8 | 1 |

(iii) The coin is tossed four times.

So sample space S = { HHHH, HHHT,HHTH, HHTT, HTHH,HTHT, HTTH HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}. n = 16

X ; number of heads in four tosses of a coin.

X = 0, 1, 2, 3, 4.

P(X = 0) = 1/16, P(X = 1) = 4/16, P(X = 2) = 6/16, P(X = 3) = 4/16, P(X = 4) = 1/16.

The probability distribution function is as follows.

X | 0 | 1 | 2 | 3 | 4 | Total |

p(X) | 1/16 | 4/16 | 6/16 | 4/16 | 1/16 | 1 |

Question (5)

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as (i) number greater than 4 (ii) six appears on at least one dieSolution

The die is thrown two times. so sample space will contain 36 elements. n = 36.(i) The success is defined as number greater than 4.

X ; the number of sucess

X = 0, 1, 2.

p = probability of seccess = 2/6

\[P\left( {x = 0} \right) = \frac{4}{6} \times \frac{4}{6} = \frac{{16}}{{36}}\] \[P\left( {x = 1} \right) = \frac{4}{6} \times \frac{2}{6} + \frac{2}{6} \times \frac{4}{6} = \frac{{8 + 8}}{{36}} = \frac{{16}}{{36}}\] \[P\left( {x = 2} \right) = \frac{2}{6} \times \frac{2}{6} = \frac{4}{{36}}\] The probability distribution function is given as follows.

X | 0 | 1 | 2 | Total |

P(X) | 16/36 | 16/36 | 4/36 | 1 |

(ii) Six appears on at least one die.

X = 0, 1.

P(X = 0) = 25/36, P(X = 1) = 11/36.

X | 0 | 1 | Total |

P(X) | 25/36 | 11/36 | 1 |

Question (6)

From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.Solution

6 bulbs out of 30 bulbs are selected with replacement.The probability of selected bulb is defective = 6/30 = 1/5.

The probability that selected bulb is not defective = 24/ 30 = 4/5.

X = Number of defective bulbs

X = 0, 1, 2, 3, 4.

\[P\left( {x = 0} \right) = \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} = \frac{{256}}{{625}}\] \[P\left( {x = 1} \right) = 4C1 \times \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} \times \frac{1}{5} = 4 \times \frac{{64}}{{125}} \times \frac{1}{5} = \frac{{256}}{{625}}\] \[P\left( {x = 2} \right) = 4{C_2} \times \frac{4}{5} \times \frac{4}{5} \times \frac{1}{5} \times \frac{1}{5} = 6 \times \frac{{16}}{{25}} \times \frac{1}{{25}} = \frac{{96}}{{625}}\] \[P\left( {x = 3} \right) = 4{C_3} \times \frac{4}{5} \times \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = 4 \times \frac{4}{5} \times \frac{1}{{125}} = \frac{{16}}{{625}}\] \[P\left( {x = 4} \right) = \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = \frac{1}{{625}}\] The probability distribution function is given as follows.

X | 0 | 1 | 2 | 3 | 4 | Total |

p(X) | 256/625 | 256/625 | 96/625 | 16/625 | 1/625 | 1 |

Question (7)

A coin is biased 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.Solution

P(H) = 3P(T)P(H) + P(T) = 1

3P(T) + P(T) = 1

4P(T) = 1

P(T) = 1/4, P(H) = 3/4.

A coin is tossed twice , so sample space S = {HH, HT, TH, TT}

X ; the number of tails.

X = 0, 1, 2.

P(X = 0 ) = P( HH)

\[P\left( {x = 0} \right) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{{16}}\] P(X = 1) = P( HT, TH)

\[P\left( {x = 1} \right) = \frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4} = \frac{{3 + 3}}{{16}} = \frac{6}{{16}}\] P(X = 2) = P(TT)

\[P\left( {x = 2} \right) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{{16}}\]The probability distribution function is as follows

X | 0 | 1 | 2 | Total |

P(X) | 9/16 | 6/16 | 1/16 | 16/16=1 |

Question (8)

. A random variable X has the following probability distribution:X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

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

Solution

(i) As it is given the probability distribution function,∑p(x) = 1

0 + k + 2k + 2k + 3k + k

10k

10 k

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

k + 1 = 0 or 10k - 1 = 0

k = -1 or k = 1/10

As k is the probability k ≠ -1. so k = 1/10.

(ii) P(X < 3) = P(0) + P(1) + P(2)

= 0 + k + 2k

=3k

= 3/10

(iii) P( X > 6) = P(7)

= 7k

= 7 (1/10)

= 7/100 + 1/10

= 17/100

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

= k + 2k

= 3k

= 3/10.

Question (9)

The random variable X has probability distribution P(X) of the following form, where k is some number: P(X) = { 𝑘, if x = 0 2𝑘, if x = 1 3𝑘, if x = 2 0, otherwise (a) Determine the value of k. (b) Find P(X < 2),P( X≤2) , P(X ≥ 2),Solution

( a ) The probability distribution function for the random variale is given.∑p(x) = 1

k + 2k + 3k +0 = 1

6k = 1

k = 1/6

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

= k + 2k

= 3k

= 3(1/6) = 1/2

P( X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= k + 2k + 3k

= 6k

=6(1/6) = 1

P(X ≥ 2) = P(X = 2)

= 3k

=3(1/6) = 1/2.

Question (10)

Find the mean number of heads in three tosses of a fair coin.Solution

As coin is toss three times , n = 8X ; the number of heads.

X = 0, 1, 2, 3.

P(X = 0) = 1/8, P(X = 1) = P(X = 2) = 3/8, P(X = 3) = 1/8.

X | P(X) | X P(X) |

0 | 1/8 | 0 |

1 | 3/8 | 3/8 |

2 | 3/8 | 6/8; |

3 | 1/8 | 3/8 |

Total | 1 | 12/8 |

Mean of X = E(X)

= ∑ XP(X)

= 12/8

= 1.5

Question (11)

Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.Solution

Two dics are thrown , so n = 36.X ; the number of sixes.

X = 0, 1, 2.

P(X = 0) = 25/36, P(X = 1) = 10/36, P(X = 2) = 1/36

X | P(X) | X P(X) |

0 | 25/36 | 0 |

1 | 10/36 | 10/36 |

2 | 1/36 | 2/36 |

Total | 36/36 = 1 | 12/36; |

E(X) = ∑XP(X)

= 12/36

= 1/3.

Question (12)

Two numbers are selected at random (without replacement) from the first six positive integers. Let X denotes the larger of the two numbers obtained. Find E(X).Solution

There are 6 positive numbers , Two number are selected with out replacement.S = {(1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (5,6)}

n = 15.

X ; the large number between them

X = 2, 3, 4, 5, 6.

X | P(X) | X P(X) |

2 | 1/15 | 2/15 |

3 | 2/15 | 6/15 |

4 | 3/15 | 12/15 |

5 | 4/15 | 20/15 |

6 | 5/15 | 30/15 |

Total | 15/15=1 | 70/15 |

E(X) = ∑XP(X)

= 70/15

= 14/3

Question (13)

Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of XSolution

When two dice are rolled then n = 36.X ; Sum og the numbers on dics.

X = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

X | P(X) | XP(X) | X^{2}P(X) |

2 | 1/36 | 2/36 | 4/36 |

3 | 2/36 | 6/36 | 18/36 |

4 | 3/36/td> | 12/36 | 48/36 |

5 | 4/36 | 20/36 | 100/36 |

6 | 5/36 | 30/36 | 180/36 |

7 | 6/36 | 42/36 | 294/36 |

8 | 5/36 | 40/36; | 320/36 |

9 | 4/36 | 36/36 | 324/36 |

10 | 3/36 | 30/36 | 300/36 |

11 | 2/36 | 22/36 | 242/36 |

12 | 1/36 | 12/36 | 144/36 |

Total | 36/36 = 1 | /252/36 | 1974/36 |

E(X) = ∑XP(X)

= 252 / 36

= 7

E(X

= 1974/ 36

=54.833

V(X) = E(X

= 54.833 - (7)

= 54.833 - 49

= 5.833

S.D = √V(X)

= √5.833

= 2.415.

Question (14)

A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.Solution

X ; age of the studentX = 14,15, 16, 17, 18, 19, 20, 21.

X | frquency | P(X) | XP(X) | X^{2}P(X) |

14 | 2 | 2/15 | 28/15 | 392/15 |

15 | 1 | 1/15 | 15/15 | 225/15 |

16 | 2 | 2/15 | 32/15 | 512/15 |

17 | 3 | 3/15 | 51/15 | 867/15 |

18 | 1 | 1/15 | 18/15 | 324/15 |

19 | 2 | 2/15 | 38/15 | 722/15 |

20 | 3 | 3/15 | 60/15 | 1200/15 |

21 | 1 | 1/15 | 21/15 | 441/15 |

Total | 15 | 1 | 263/15 | 4683/15 |

E(X) = ∑XP(X)

= 263/15

=17.53

E(X

= 4683/15

V(X) = E(X

= 4683/ 15 - (263/15)

= (70245 - 69169)/225

= 1076 / 225

=4.78

S.D.(X) = √V(X)

= √4.78

=2.186 = 2.19.

Question (15)

. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var(X).Solution

IN a meeting 70% favors and 30% oppose a proposal.X = 0 for oppose and X = 1 for favour.

X | P(X) | XP(X) | X^{2}P(X) |

0 | 0.3 | 0 | 0 |

1 | 0.7 | 0.7 | 0.7 |

Total | 1 | 0.7 | 0.7 |

E(X) = ∑XP(X)

= 0.7

E(X

= 0.7

V(X) = E(X

= 0.7 - (0.7)

= 0.7 - 0.49

= 0.21

Question (16)

The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is(A) 1 (B) 2 (C) 5 (D) 8

Solution

X = 1, 2, 5.X | P(X) | XP(X) |

1 | 3/6 | 3/6 |

2 | 2/6 | 4/6 |

5 | 1/6 | 5/6 |

Total | 1 | 12/6 |

Mean E(X) = ∑XP(X)

=12/6

= 2

So B is the correct answer.

Question (17)

Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is(A) 37/221 (B) 5/13 (C) 1/13 (D) 2/13

Solution

Two cards are drawn from a pack of cards.the probability of ace card = 4/52 = 1/13

X ; the number of ace card

X = 0, 1, 2.

X | P(X) | XP(X) |

0 | 144/169 | 0 |

1 | 24/169 | 24/169 |

2 | 1/169 | 2/169 |

Total | 1 | 26/169 |

E(X) = ∑XP(X)

= 26/169

= 2/13

D is the correct answer.