NCERT Solutions for Exercise 13.4 Class 12 Maths Chapter 13 - Probability

Question:1(ii) State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

Answer:

As we know probabilities cannot be negative for a probability distribution .

$P(3) = -0.1$

$\therefore$ The given table is not a the probability distributions of a random variable.

Question:1(iii) State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

Answer:

As we know sum of probabilities of a probability distribution is 1.

Sum of probablities $=0.6+0.1+0.2=0.9\neq 1$

$\therefore$ The given table is not a the probability distributions of a random variable because sum of probabilities is not 1.

Question:1(iv) State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

Answer:

As we know sum of probabilities of a probability distribution is 1.

Sum of probablities $=0.3+0.2+0.4+0.1+0.05=1.05\neq 1$

$\therefore$ The given table is not a the probability distributions of a random variable because sum of probabilities is not 1.

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 ?

Answer:

B = black balls

R = red balls

The two balls can be selected as BR,BB,RB,RR.

X = number of black balls.

$\therefore$ $X(BB)=2$

$X(RB)=1$

$X(BR)=1$

$X(RR)=0$

Hence, possible values of X can be 0, 1 and 2.

Yes, X is a random variable.

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 possibl valuess of $X$?

Answer:

The difference between the number of heads and the number of tails obtained when a coin is tossed $6$ times are :

$\therefore$ $X(6H,0T)=\left | 6-0 \right |=6$

$\Rightarrow \, \, X(5H,1T)=\left | 5-1 \right |=4$

$\Rightarrow \, \, X(4H,2T)=\left | 4-2 \right |=2$

$\Rightarrow \, \, X(3H,3T)=\left | 3-3 \right |=0$

$\Rightarrow \, \, X(2H,4T)=\left | 2-4\right |=2$

$\Rightarrow \, \, X(1H,5T)=\left | 1-5\right |=4$

$\Rightarrow \, \, X(0H,6T)=\left |0-6\right |=6$

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

Question:4(i) Find the probability distribution of

number of heads in two tosses of a coin.

Answer:

When coin is tossed twice then sample space $=\left \{ HH,HT,TH,TT \right \}$

Let X be number of heads.

$\therefore$ $X(HH)=2$

$X(HT)=1$

$X(TH)=1$

$X(TT)=0$

X can take values of 0,1,2.

$P(HH)=P(HT)=P(TH)=P(TT)=\frac{1}{4}$

$P(X=0)=P(TT)=\frac{1}{4}$

$P(X=2)=P(HH)=\frac{1}{4}$

$P(X=1)=P(HT)+P(TH)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Table is as shown :

Question:4(ii) Find the probability distribution of

number of tails in the simultaneous tosses of three coins.

Answer:

When 3 coins are simultaneous tossed then sample space $=\left \{ HHH,HHT,TTH,TTT,HTH,THT,THH,HTT \right \}$

Let X be number of tails.

$\therefore$ X can be 0,1,2,3

X can take values of 0,1,2.

$P(X=0)=P(HHH)=\frac{1}{8}$

$P(X=1)=P(HHT)+P(HTH)+P(THH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

$P(X=2)=P(THT)+P(HTT)+P(TTH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

$P(X=3)=P(TTT)=\frac{1}{8}$

Table is as shown :

Question:4(iii) Find the probability distribution of

number of heads in four tosses of a coin.

Answer:

When coin is tossed 4 times then sample space $=\left \{ HHHH,HHHT,TTTH,HTHH,THHT,TTHH,HHTT ,TTTT,HTTH,THTH,HTHT,HTTT,THHH,THTT,HHTH,TTHT\right \}$

Let X be number of heads.

$\therefore$ X can be 0,1,2,3,4

$P(X=0)=P(TTTT)=\frac{1}{16}$

$P(X=1)=P(HTTT)+P(TTTH)+P(THTT)=(TTHT)=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}$

$P(X=2)=P(THHT)+P(HHTT)+P(HTTH)+P(TTHH)+P(HTHT)+P(THTH)=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{6}{16}=\frac{3}{8}$

$P(X=3)=P(HHHT)+P(THHH)+P(HHTH)+P(HTHH)=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}$

$P(X=4)=P(HHHH)=\frac{1}{16}$

Table is as shown :

Question:5(i) Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as

number greater than 4

Answer:

When a die is tossed twice , total outcomes = 36

Number less than or equal to 4 in both toss : $P(X=0)=\frac{4}{6} \times \frac{4}{6}=\frac{4}{9}$

Number less than or equal to 4 in first toss and number more than or equal to 4 in second toss + Number less than or equal to 4 in second toss and number more than or equal to 4 in first toss: $P(X=1)=\frac{4}{6} \times \frac{2}{6}+ \frac{2}{6}\times \frac{4}{6}=\frac{4}{9}$

Number less than 4 in both tosses : $P(X=2)=\frac{2}{6} \times \frac{2}{6}= \frac{1}{9}$

Probability distribution is as :

Question:5(ii) Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as

six appears on at least one die.

Answer:

When a die is tossed twice , total outcomes = 36

Six does not appear on any of the die : $P(X=0)=\frac{5}{6} \times \frac{5}{6}=\frac{25}{36}$

Six appear on atleast one die : $P(X=1)=\frac{1}{6} \times \frac{5}{6}+ \frac{1}{6}\times \frac{5}{6}=\frac{5}{18}$

Probability distribution is as :

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.

Answer:

Total bulbs = 30

defective bulbs = 6

Non defective bulbs $=30-6=24$

$P(defective \, \, \, \, \, bulbs)=\frac{6}{30}=\frac{1}{5}$

$P(Non \, \, \, \, defective \, \, \, \, \, bulbs)=\frac{24}{30}=\frac{4}{5}$

$4$ bulbs is drawn at random with replacement.

Let X : number of defective bulbs

4 Non defective bulbs and 0 defective bulbs : $P(X=0)={4}\textrm{C}_0\frac{4}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}$

3 Non defective bulbs and 1 defective bulbs : $P(X=1)={4}\textrm{C}_1\frac{1}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}$

2 Non defective bulbs and 2 defective bulbs : $P(X=2)={4}\textrm{C}_2\frac{1}{5}.\frac{1}{5}.\frac{4}{5}.\frac{4}{5}=\frac{96}{625}$

1 Non defective bulbs and 3 defective bulbs : $P(X=3)={4}\textrm{C}_3\frac{1}{5}.\frac{1}{5}.\frac{1}{5}.\frac{4}{5}=\frac{16}{625}$

0 Non defective bulbs and 4 defective bulbs : $P(X=4)={4}\textrm{C}_4\frac{1}{5}.\frac{1}{5}.\frac{1}{5}.\frac{1}{5}=\frac{1}{625}$

the probability distribution of the number of defective bulbs is as :

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.

Answer:

the coin is tossed twice, total outcomes =4 $=\left \{ HH,TT,HT,TH \right \}$

probability of getting a tail be x.

i.e. $P(T)=x$

Then $P(H)=3x$

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

$4x=1$

$x=\frac{1}{4}$

$P(T)=\frac{1}{4}$ and $P(H)=\frac{3}{4}$

Let X : number of tails

No tail : $P(X=0)=P(H).P(H)=\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}$

1 tail : $P(X=1)=P(HT)+P(TH)=\frac{3}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{3}{4}=\frac{3}{8}$

2 tail : $P(X=2)=P(TT)=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$

the probability distribution of number of tails are

Question:8(i) A random variable X has the following probability distribution:

$k$

Answer:

Sum of probabilities of probability distribution of random variable is 1.

$\therefore \, \, \, \, 0+k+2k+2k+3k+k^{2}+2k^{2}+7k^{2}+k=1$

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

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

$k=\frac{1}{10}\, \, and\, \, k=-1$

Question:8(ii) A random variable $X$ has the following probability distribution:

$P(X< 3)$

Answer:

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

$=0+K+2K$

$=3K$

$=3\times \frac{1}{10}$

$= \frac{3}{10}$

Question:8(iii) A random variable $X$ has the following probability distribution:

$P(X> 6)$

Answer:

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

$=7K^2+K$

$=7\times (\frac{1}{10})^2+\frac{1}{10}$

$=\frac{7}{100}+\frac{1}{10}$

$=\frac{17}{100}$

Question:8(iv) A random variable X has the following probability distribution:

$P(0< X< 3)$

Answer:

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

$=k+2k$

$=3k$

$=3\times \frac{1}{10}$

$= \frac{3}{10}$

Question:9(a) The random variable X has a probability distribution P(X) of the following form, where k is some number :

$P(X)=\left\{\begin{matrix} k, if&i x=0\\ 2k, if& x=1\\ 3k, if& x=2\\ 0,& otherwise \end{matrix}\right.$

Determine the value of $k.$

Answer:

Sum of probabilities of probability distribution of random variable is 1.

$\therefore \, \, \, \, k+2k+3k+0=1$

$6k=1$

$k=\frac{1}{6}$

Question:9(b) The random variable $X$ has a probability distribution $P(X)$ of the following form, where k is some number :

$P(X)=\left\{\begin{matrix} k,& if & x=0\\ 2k,& if& x=1\\ 3k,& if& x=2\\ 0,& otherwise& \end{matrix}\right.$

Find $P\left ( X< 2 \right ),P\left ( X\leq 2 \right ),P\left ( X\geq 2 \right ).$

Answer:

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

$=k+2k$

$=3k$

$=3\times \frac{1}{6}$

$= \frac{1}{2}$

$P(X\leq 2)=P(X=0)+P(X=1)+p(X=2)$

$=k+2k+3k$

$=6k$

$=6\times \frac{1}{6}$

$=1$

$P(X\geq 2)=P(X=2)+P(X> 2)$

$=3k+0$

$=3\times \frac{1}{6}=\frac{1}{2}$

Question:10 Find the mean number of heads in three tosses of a fair coin.

Answer:

Let X be the success of getting head.

When 3 coins are tossed then sample space $=\left \{ HHH,HHT,TTH,TTT,HTH,THT,THH,HTT \right \}$

$\therefore$ X can be 0,1,2,3

$P(X=0)=P(TTT)=\frac{1}{8}$

$P(X=1)=P(HTT)+P(TTH)+P(HTH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

$P(X=2)=P(HHT)+P(HTH)+P(THH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

$P(X=3)=P(HHH)=\frac{1}{8}$

The probability distribution is as

mean number of heads :

$=0\times \frac{1}{8}+1\times \frac{3}{8}+2\times \frac{3}{8}+3\times \frac{1}{8}$

$= \frac{3}{8}+ \frac{6}{8}+ \frac{3}{8}$

$=\frac{12}{8}$

$=\frac{3}{2}=1.5$

Question:11 Two dice are thrown simultaneously. If $X$ denotes the number of sixes, find the expectation of $X$.

Answer:

$X$ denotes the number of sixes, when two dice are thrown simultaneously.

X can be 0,1,2.

$\therefore$ Not getting six on dice $P(X)=\frac{25}{36}$

Getting six on one time when thrown twice : $P(X=1)=2\times \frac{5}{6}\times \frac{1}{6}=\frac{10}{36}$

Getting six on both dice : $P(X=2)= \frac{1}{36}=\frac{1}{36}$

Expectation of X = E(X)

$E(X)=0\times \frac{25}{36}+1\times \frac{10}{36}+2\times \frac{1}{36}$

$E(X)= \frac{12}{36}$

$E(X)= \frac{1}{3}$

Question:12 Two numbers are selected at random (without replacement) from the first six positive integers. Let $X$ denote the larger of the two numbers obtained. Find $E(X).$

Answer:

Two numbers are selected at random (without replacement) from the first six positive integers in $6\times 5=30$ ways.

$X$ denote the larger of the two numbers obtained.

X can be 2,3,4,5,6.

X=2, obsevations : $(1,2),(2,1)$

$P(X=2)=\frac{2}{30}=\frac{1}{15}$

X=3, obsevations : $(1,3),(3,1),(2,3),(3,2)$

$P(X=3)=\frac{4}{30}=\frac{2}{15}$

X=4, obsevations : $(1,4),(4,1),(2,4),(4,2),(3,4),(4,3)$

$P(X=4)=\frac{6}{30}=\frac{3}{15}$

X=5, obsevations : $(1,5),(5,1),(2,5),(5,2),(3,5),(5,3),(4,5),(5,4)$

$P(X=5)=\frac{8}{30}=\frac{4}{15}$

X=6, obsevations : $(1,6),(6,1),(2,6),(6,2),(3,6),(6,3),(4,6),(6,4),(5,6),(6,5)$

$P(X=6)=\frac{10}{30}=\frac{1}{3}$

Probability distribution is as follows:

$E(X)=2\times \frac{1}{15}+3\times \frac{2}{15}+4\times \frac{3}{15}+5\times \frac{4}{15}+6\times \frac{1}{3}$

$E(X)= \frac{2}{15}+ \frac{2}{5}+ \frac{4}{5}+ \frac{4}{3}+ \frac{2}{1}$

$E(X)= \frac{70}{15}$

$E(X)= \frac{14}{3}$

Question:13 Let $X$ denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of$X$.

Answer:

$X$ denote the sum of the numbers obtained when two fair dice are rolled.

Total observations = 36

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

$P(X=2)=P(1,1)=\frac{1}{36}$

$P(X=3)=P(1,2)+P(2,1)=\frac{2}{36}=\frac{1}{18}$

$P(X=4)=P(1,3)+P(3,1)+P(2,2)=\frac{3}{36}=\frac{1}{12}$

$P(X=5)=P(1,4)+P(4,1)+P(2,3)+P(3,2)=\frac{4}{36}=\frac{1}{9}$

$P(X=6)=P(1,5)+P(5,1)+P(2,4)+P(4,2)+P(3,3)=\frac{5}{36}$

$P(X=7)=P(1,6)+P(6,1)+P(2,5)+P(5,2)+P(3,4)+P(4,3)=\frac{6}{36}=\frac{1}{6}$

$P(X=8)=P(2,6)+P(6,2)+P(3,5)+P(5,3)+P(4,4)=\frac{5}{36}$$P(X=9)=P(3,6)+P(6,3)+P(4,5)+P(5,4)=\frac{4}{36}=\frac{1}{9}$

$P(X=10)=P(4,6)+P(6,4)+P(5,5)=\frac{3}{36}=\frac{1}{12}$

$P(X=11)=P(5,6)+P(6,5)=\frac{2}{36}=\frac{1}{18}$

$P(X=12)=P(6,6)=\frac{1}{36}$

Probability distribution is as follows :

$E(X)=2\times \frac{1}{36}+3\times \frac{1}{18}+4\times \frac{1}{12}+5\times \frac{1}{9}+6\times \frac{5}{36}+7\times \frac{1}{6}+8\times \frac{5}{36}+9\times \frac{1}{9}+10\times \frac{1}{12}+11\times \frac{1}{18}+12\times \frac{1}{36}$

$E(X)=\frac{1}{18}+\frac{1}{6}+\frac{1}{3}+\frac{5}{9}+\frac{5}{6}+\frac{7}{6}+\frac{10}{9}+1+\frac{5}{6}+\frac{11}{18}+\frac{1}{3}$

$E(X)=7$

$E(X^2)=4\times \frac{1}{36}+9\times \frac{1}{18}+16\times \frac{1}{12}+25\times \frac{1}{9}+36\times \frac{5}{36}+49\times \frac{1}{6}+64\times \frac{5}{36}+81\times \frac{1}{9}+100\times \frac{1}{12}+121\times \frac{1}{18}+144\times \frac{1}{36}$

$E(X^2)=\frac{987}{18}=\frac{329}{6}=54.833$

$Variance = E(X^2)-(E(X))^2$

$=54.833-7^2$

$=54.833-49$

$=5.833$

Standard deviation =$=\sqrt{5.833}=2.415$

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 $X$.

Answer:

Total students = 15

probability of selecting a student :

$=\frac{1}{15}$

The information given can be represented as frequency table :

$P(X=14)=\frac{2}{15}$ $P(X=15)=\frac{1}{15}$ $P(X=16)=\frac{2}{15}$

$P(X=17)=\frac{3}{15}=\frac{1}{5}$ $P(X=18)=\frac{1}{15}$ $P(X=19)=\frac{2}{15}$

$P(X=20)=\frac{3}{15}=\frac{1}{5}$ $P(X=21)=\frac{1}{15}$

Probability distribution is as :

$E(X)=14\times \frac{2}{15}+15\times \frac{1}{15}+16\times \frac{2}{15}+17\times \frac{1}{5}+18\times \frac{1}{15}+19\times \frac{2}{15}+20\times \frac{1}{5}+21\times \frac{1}{15}$

$E(X)=\frac{263}{15}=17.53$

$E(X^2)=14^2\times \frac{2}{15}+15^2\times \frac{1}{15}+16^2\times \frac{2}{15}+17^2\times \frac{1}{5}+18^2\times \frac{1}{15}+19^2\times \frac{2}{15}+20^2\times \frac{1}{5}+21^2\times \frac{1}{15}$

$E(X^2)=\frac{4683}{15}=312.2$

$Variance =E(X^2)-(E(X))^2$

$Variance =312.2-(17.53)^2$

$Variance =312.2-307.42$

$Variance =4.78$

$Standard\, \, deviation =\sqrt{4.78}=2.19$

Question:15 In a meeting, $70^{o}/_{o}$ of the members favour and $30^{o}/_{o}$ 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 $(X)$.

Answer:

Given :

$P(X=0)=30\%=\frac{30}{100}=0.3$

$P(X=1)=70\%=\frac{70}{100}=0.7$

Probability distribution is as :

$E(X)=0\times 0.3+1\times 0.7$

$E(X)= 0.7$

$E(X^2)=0^2\times 0.3+1^2\times 0.7$

$E(X^2)= 0.7$

$Variance = E(X^2)-(E(X))^2$

$Variance = 0.7-(0.7)^2$

$Variance = 0.7-0.49=0.21$

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, Choose the correct answer in the following:

(A) $1$

(B) $2$

(C) $5$

(D) $\frac{8}{3}$

Answer:

X is number representing on die.

Total observations = 6

$P(X=1)=\frac{3}{6}=\frac{1}{2}$

$P(X=2)=\frac{2}{6}=\frac{1}{3}$

$P(X=5)=\frac{1}{6}$

$E(X)=1\times \frac{1}{2}+2\times \frac{1}{3}+5\times \frac{1}{6}$

$E(X)= \frac{1}{2}+ \frac{2}{3}+ \frac{5}{6}$

$E(X)=\frac{12}{6}=2$

Option B is correct.

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 Choose the correct answer in the following:

(A) $\frac{37}{221}$

(B) $\frac{5}{13}$

(C) $\frac{1}{13}$

(D) $\frac{2}{13}$

Answer:

X be number od aces obtained.

X can be 0,1,2

There 52 cards and 4 aces, 48 are non-ace cards.

$P(X=0)=P(0 \, ace\, and\, 2\, non\, ace\, cards)=\frac{^4C_2.^4^8C_2}{^5^2C_2}=\frac{1128}{1326}$

$P(X=1)=P(1 \, ace\, and\, 1\, non\, ace\, cards)=\frac{^4C_1.^4^8C_1}{^5^2C_2}=\frac{192}{1326}$

$P(X=2)=P(2 \, ace\, and\, 0\, non\, ace\, cards)=\frac{^4C_2.^4^8C_0}{^5^2C_2}=\frac{6}{1326}$

The probability distribution is as :

$E(X)=0\times \frac{1128}{1326}+1\times \frac{192}{1326}+2\times \frac{6}{1326}$

$E(X)=\frac{204}{1326}$

$E(X)=\frac{2}{13}$

Option D is correct.