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Probability · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 28 questions.

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Question 11 Mark
Two cards are drawn at random and one-by-one without replacement from a well-shuffled pack of 52 playing cards. Find the probability that one card is red and the other is black.
Answer
Given,
R = 26
B = 26
Required Probability $=\frac{26}{52}\times\frac{26}{51}+\frac{26}{52}\times\frac{26}{51}$
$=2\times\Big(\frac{26}{52}\times\frac{26}{51}\Big)$
$=\frac{26}{51}$
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Question 21 Mark
Compute $\text{P}(\text{A}|\text{B})\ \text{if}\ \text{P}(\text{B})=0.5\ \text{and}\ \text{P}(\text{A}\cap\text{B})=0.32$ 
Answer
Given:    $\text{P}\left(\text{B}\right)=0.5,\ \text{P}\left(\text{A}\cap\text{B}\right)=0.32$$\therefore\ \ \ \ \ \ \ \ \ \ \ \text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{0.32}{0.50}=\frac{32}{50}=0.64$
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Question 31 Mark
If $\text{P}(\text{A})=\frac{3}{5}\text{and}\ \text{P}(\text{B})=\frac{1}{5}$, find $\text{P}(\text{A}\cap\text{B})$ if A and B are independent events.
Answer
It is given that $\text{P}(\text{A})=\frac{3}{5}\ \text{and}\ \text{P}(\text{B})=\frac{1}{5}$A and B are independent events. Therefore,
$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A}).\text{P}(\text{B})=\frac{3}{5}.\frac{1}{5}=\frac{3}{25}$
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Question 41 Mark
Evaluate $\text{P}(\text{A}\cap\text{B})\ \text{if}\ 2\text{P}(\text{A})=\text{P}(\text{B})=\frac{5}{13}\ \text{and}\ \text{P}(\text{A}|\text{B})=\frac{2}{5}.$
Answer
$\text{Given:}\ \ \ \ \ 2\text{P}\left(\text {A}\right)=\text{P}\left(\text{B}\right)=\frac{5}{13},\ \text{P}\left(\text{A}|\text{B}\right)=\frac{2}{5}$
$ \therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \text{P}\left(\text{A}\right)=\frac{5}{26}$
$\text{Now}\ \ \ \ \ \ \ \ \ \text{P}\left (\text{A}\cap\text{B}\right)=\text{P}\left(\text{B}|\text{A}\right).\text{P}\left(\text{B}\right)=\frac{2}{5}\times\frac{5}{13}=\frac {2}{13}$
$\text{And}\ \ \ \ \ \ \ \ \ \ \text{P}\left (\text{A}\cup\text{B}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)-\text{P}\left(\text{A}\cap\text{B}\right)= \frac{5}{26}+\frac{5}{13}-\frac{2}{13}=\frac{11}{26}$
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Question 51 Mark
The probability distribution of random variable X is given below:
$\text{X}$
$0$
$1$
$2$
$3$
$\text{P}(\text{X})$
$\text{k}$
$\frac{\text{k}}{2}$
$\frac{\text{k}}{4}$
$\frac{\text{k}}{8}$
 Find $\text{P}(\text{X}\leq2)+\text{P}(\text{X}>2)$
Answer
$\text{P}(\text{X}\leq2)+\text{P}(\text{X}>2)$
$=\text{P}(0)+\text{P}(1)+\text{P}(2)+\text{P}(3)$
$=1$
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Question 61 Mark
If A and B are two events such that $\text{P}(\text{A})=\frac{1}{4},\ \text{P}(\text{B})=\frac{1}{2}\ \text{and}\ \text{P}(\text{A}\cap\text{B})=\frac{1}{8},$ find P(not A and not B).
Answer
It is given that, $\text{P}(\text{A})=\frac{1}{2}\ \text{and}\ \text{P}(\text{A}\cap\text{B})=\frac{1}{8}$P(not on A and not on B) $=\text{P}(\text{A}'\cap\text{B}')$
P(not on A and not on B) $=\text{P}((\text{A}\cup\text{B}))'\left[\text{A}'\cap\text{B}'=(\text{A}\cup\text{B})'\right]$
$=1-\text{P}(\text{A}\cup\text{B})$
$=1-\left[\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})\right]$
$=1-\Bigg[\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\Bigg]$
$=1-\frac{5}{8}$
$=\frac{3}{8}$
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Question 71 Mark
Events A and B are such that $\text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{B})=\frac{7}{12}\ \text{and}\ \text{P}(\text{not A or not B})=\frac{1}{4}.$ State whether A and B are independent?
Answer
It is given that $\text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{B})=\frac{7}{12},\ \text{and}\ \text{P}(\text{not A or not B})=\frac{1}{4}$$\Rightarrow\text{P}(\text{A}'\cup\text{B}')=\frac{1}{4}$
$\Rightarrow\text{P}\Big((\text{A}\cap\text{B})'\Big)=\frac{1}{4}\ \ \ \ \ \Big[\text{A}'\cup\text{B}'=(\text{A}\cap\text{B})'\Big]$
$\Rightarrow1-\text{P}(\text{A}\cap\text{B})=\frac{1}{4}$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=\frac{3}{4}\ \ \ ...(1)$ 
$\text{However},\ \text{P}(\text{A})\cdot\text{P}(\text{B})=\frac{1}{2}\cdot\frac{7}{12}=\frac{7}{24}\ \ \ ...(2)$
$\text{Here},\ \frac{3}{4}\neq\frac{7}{24}$
$\therefore\text{P}(\text{A}\cap\text{B})\neq\text{P}(\text{A})\cdot\text{P}(\text{B})$
Therefore, A and B are independent events.
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Question 81 Mark
There are 5 cards numbered 1 to 5, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on two cards drawn. Find the mean and variance of X.
Answer
Here, S = {(1,2), (2, 1), (1,3), (3, 1), (2, 3), (3, 2), (1,4), (4, 1), (1,5), (5. 1), (2, 4), (4, 2), (2, 5), (5, 2), (3, 4), (4, 3), (3, 5), (5, 3), (5, 4), (4, 5)}
⇒ n(S) = 20
Let random variable be X which denotes the sum of the numbers on two cards drawn.
$\therefore\text{X}=3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9$
At $\text{X}=3,\text{P(x)}=\frac{2}{20}=\frac{1}{10}$
At $\text{X}=4,\text{P(x)}=\frac{2}{20}=\frac{1}{10}$
At $\text{X}=5,\text{P(x)}=\frac{4}{20}=\frac{1}{5}$
At $\text{X}=6,\text{P(x)}=\frac{4}{20}=\frac{1}{5}$
At $\text{X}=7,\text{P(x)}=\frac{4}{20}=\frac{1}{5}$
At $\text{X}=8,\text{P(x)}=\frac{2}{20}=\frac{1}{10}$
At $\text{X}=9=,\text{P}(\text{X})=\frac{2}{20}=\frac{1}{10}$
$\therefore\text{Mean}, \text{E}(\text{X})=\sum\text{XP}(\text{X})$
$=\frac{3}{10}+\frac{4}{10}+\frac{5}{5}+\frac{6}{5}+\frac{7}{5}+\frac{8}{10}+\frac{9}{10}$
$=\frac{3+4+10+12+14+8+9}{10}=6$
Also, $\sum\text{X}^2\text{P}(\text{X})=\frac{9}{10}+\frac{16}{10}+\frac{25}{5}+\frac{36}{5}+\frac{49}{5}+\frac{64}{10}+\frac{81}{10}$
$=\frac{9+16+50+72+98+64+81}{10}=39$
$\therefore\text{Var}(\text{X})=\sum\text{X}^{2}\text{P}(\text{X})-\Big[\sum\text{XP}(\text{X})\Big]^2$
$=39-(6)^2=39-36=3$
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Question 91 Mark
State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
X 0 1 2
P(X) 0.4 0.4 0.2
Answer
$\text{P}(0)+\text{P}(1)+\text{P}(2)=0.4+0.4+0.2=1$
Therefore, it is a probability distribution.
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Question 101 Mark
State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
Z 3 2 1 0 -1
P(Z) 0.3 0.2 0.4 0.1 0.05
Answer
$\text{P}(3)+\text{P}(2)+\text{P}(1)+\text{P}(0)+\text{P}(-1)=0.3+0.2+0.4+0.1+0.05=1.05\neq1$
Therefore, it is not a probability distribution.
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Question 111 Mark
A fair coin is tossed 8 times, find the probability of.
exactly 5 heads.
Answer
Let X denote the number of heads obtained when a fair is tossed 8 times.
Now, X is a binomial distribution with $\text{n}=8,\text{p}=\frac{1}{2}$ and $\text{q}=1-\frac{1}{2}=\frac{1}{2}.$
$\therefore\text{P}(\text{X = r})=\text{ }^8\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{8-\text{r}}=\text{ }^8\text{C}_{\text{r}}\big(\frac{1}{2}\big)^8,\text{r}=0,1,2,\dots,8$
(1) Probability of getting exactly 5 heads $=\text{P}(\text{X}=5)=\text{ }^8\text{C}_5\big(\frac{1}{2}\big)^8=\frac{56}{256}=\frac{7}{32}$
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Question 121 Mark
Given two independent events A and B such that $\text{P}(\text{A})=0.3,\ \text{P}(\text{B})=0.6.\ \text{Find}$ $\text{P}(\text{A}\ \text{and not}\ \text{B})$ 
Answer
It is given that $\text{P} (\text{A})=0.3\ \text{and}\ \text{P}(\text{B})=0.6$
Also, A and B are independent events.
P(A and not B) $=\text{P}(\text{A}\cap\text{B}')$
$ =\text{P}(\text{A})-\text{P}(\text{A}\cap\text{B})$
= 0.3 - 0.18
= 0.12
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Question 131 Mark
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
both balls are red.
Answer
Total number of balls = 18
Number of red balls = 8
Number of black balls = 10
Probability of getting a red ball in the first draw $=\frac{8}{18}=\frac{4}{9}$
The ball is replaced after the first draw.
$​​​​​​\therefore\ $Probability of getting a red ball in the second draw $=\frac{8}{18}=\frac{4}{9}$
Therefore, probability of getting both the balls red $=\frac{4}{9}\times\frac{4}{9}=\frac{16}{81}$
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Question 141 Mark
If A and B are two independent evets, then write $\text{P}(\text{A}\cap\overline{\text{B}})$ in terms of P(A) and P(B).
Answer
A and B are two independent events
$\therefore\ \text{P}(\text{A}\cap\overline{\text{B}})=\text{P(A)}\text{ P}(\overline{\text{B}})$
$=\text{P(A)}[1-\text{P(B)}]$
$=\text{P(A)}-\text{P(A)}\text{P(B)}$
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Question 151 Mark
$\text{If}\ \text{P}(\text{A})=\frac{6}{11},\ \text{P}(\text{B})=\frac{5}{11}\ \text{and}\ \text{P}(\text{A}\cup\text{B})=\frac{7}{11},$find
$\text{P}(\text{A}\cap\text{B})$
Answer
$\text{P}\left(\text{A}\cup\text{B}\right)= \text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)-\text{P}\left(\text{A}\cap\text{B}\right)$
$\Rightarrow\ \ \ \frac {7}{11}=\frac{6}{11}+\frac{5}{11}-\text{P}\left(\text{A}\cap\text{B}\right)$
$ \Rightarrow\ \ \ \text {P}\left(\text{A}\cap\text{B}\right)=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}=\frac{4}{11}$
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Question 161 Mark
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find.
$\text{P}(\text{A}|\text{B})$
Answer
It is given that P(A) = 0.3 and P(B) = 0.4
It is known that, $\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$\Rightarrow\text{P}(\text{A}|\text{B})=\frac{0.12}{0.4}=0.3$
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Question 171 Mark
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
first ball is black and second is red.
Answer
Total number of balls = 18 Number of red balls = 8 Number of black balls = 10Probability of getting first ball black $=\frac{10}{18}=\frac{5}{9}$
The ball is replaced after the first draw.
Probability of getting second ball as red $=\frac{8}{18}=\frac{4}{9}$
Therefore, probability of getting first ball as black and second ball as red $=\frac{5}{9}\times\frac{4}{9}=\frac{20}{81}$
 
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Question 181 Mark
State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
Y -1 0 1
P(Y) 0.6 0.1 0.2
Answer
$\text{P}(-1)+\text{P}(0)+\text{P}(1)=0.6+0.1+0.2=0.9\neq1$
Therefore, it is not a probability distribution.
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Question 191 Mark
Given that E and F are events such that $\text{P}(\text{E})=0.6,\ \text{P}(\text{F})=0.3\ \text{and}\ \text{P}(\text{E}\cap\text{F})=0.02,\ \text{find}\ \text{P}(\text{E}|\text{F})\ \text{and}\ \text{P}(\text{F}|\text{E}) .$
Answer
$\text{Given:}\ \ \text{P}\left(\text {E}\right)=0.6,\ \text{P}\left(\text{F}\right)=0.3\ \text{and P}\left(\text{E}\cap\text{F}\right)=0.2$
$\therefore\ \ \ \ \ \ \text{P}\left(\text{E|F}\right)=\frac {\text{P}\left(\text{E} \cap\text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{0.2}{0.3}=\frac{2}{3}$
$\text{And}\ \ \ \text{P}\left(\text{F}|\text {E}\right)=\frac{\text{P}\left(\text{E}\cap\text{F}\right)}{\text{P}\left(\text{E}\right)}=\frac{0.2}{0.6}=\frac{1}{3}$
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Question 201 Mark
$\text{If}\ \text{P}(\text{A})=0.8,\ \text{P}(\text{B})=0.5\ \text{and}\ \text{P}(\text{B}|\text{A})=0.4,\ \text{find}:$
$ \text{P}(\text{A}|\text{B})$
Answer
$\text{P}\left(\text{A}|\text {B}\right)=\frac{\text{P}\left(\text{A}\cap\text{B}\right)}{\text{P}\left(\text{B}\right)}=\frac{0.32}{0.50}=\frac{32}{50}=0.64$
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Question 211 Mark
$\text{If}\ \text{P}(\text{A})=\frac{6}{11},\ \text{P}(\text{B})=\frac{5}{11}\ \text{and}\ \text{P}(\text{A}\cup\text{B})=\frac{7}{11},$find
$\text{P}(\text{B}|\text{A})$
Answer
$\text{P}\left(\text{B}|\text{A}\right)= \frac{\text{P}\left(\text{A}\cap\text{B}\right)}{\text{P}\left(\text{A}\right)}=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{4}{6}=\frac{2}{3}$
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Question 221 Mark
$\text{If}\ \text{P}(\text{A})=0.8,\ \text{P}(\text{B})=0.5\ \text{and}\ \text{P}(\text{B}|\text{A})=0.4,\ \text{find}:$
$\text{P}(\text{A}\cap\text{B})$
Answer
$\text{P}(\text{A}\cap\text{B})= \text{P}(\text{B}|\text{A}).\text{P}(\text{A})=0.4\times0.8=0.32$​​​​​​​
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Question 231 Mark
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find.
$\text{P}\ (\text{A}\cap\text{B})$
Answer
It is given that P(A) = 0.3 and P(B) = 0.4
If A and B are independent events, then
$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})\cdot\text{P}(\text{B})=0.3\times0.4=0.12$
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Question 241 Mark
State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
X 0 1 2 3 4
P(X) 0.1 0.5 0.2 -1 0.3
Answer
P (3) = –0.1 which is not possible.
Therefore, it is not a probability distribution.
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Question 251 Mark
If A, B, C are mutually exclusive and exhaustive events associated to a random experiment, then write the value of P(A) + P(B) + P(C).
Answer
A, B and C are mutually exclusive and exhaustive events.
$\therefore$ P(A) + P(B) + P(C) = 1
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Question 261 Mark
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find.$\text{P}\ (\text{A}\cup\text{B})$
 
Answer
It is given that P(A) = 0.3 and P(B) = 0.4
It is known that, $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=0.3+0.4-0.12=0.58$
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Question 271 Mark
$\text{If}\ \text{P}(\text{A})=0.8,\ \text{P}(\text{B})=0.5\ \text{and}\ \text{P}(\text{B}|\text{A})=0.4,\ \text{find}:$
$\text{P}(\text{A}\cup\text{B})$
Answer
$\text{P}\left(\text{A}\cup\text {B}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)-\text{P}\left(\text{A}\cap\text{B}\right)=0.8+0.5-0.32=0.98$ 
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Question 281 Mark
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).
Answer
$\text{x}_i$ $\text{p}_i$ $\text{p}_i\text{x}_i$ $\text{p}_i\text{x}^2_i$
$0$
$1$
 		 $\frac{30}{100}$
$\frac{70}{100}$		 $0$
$\frac{70}{100}$		 $0$
$\frac{70}{100}$
    $\sum\text{p}_i\text{x}_i=\frac{70}{100}$ $\sum\text{p}_i\text{x}^2_i=\frac{70}{100}$
$\text{E}(\text{X})=\text{Mean}=\sum\text{p}_i\text{x}_i=\frac{70}{100}=0.7$
$\text{Variance}(\text{X})=\sum\text{p}_i\text{x}^2_i-\Big(\sum\text{p}_i\text{x}_i\Big)^2=\frac{70}{100}-\Bigg(\frac{70}{100}\Bigg)^2=\frac{7}{10}-\frac{49}{100}=\frac{21}{100}=0.21$
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