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Maths 2 Marks

Probability · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 117 questions.

Questions · Page 3 of 3

2 Marks

Question 1012 Marks
If A and B are two events such that $\text{P}(\text{A}\cap\text{B})=0.32$ and P (B) = 0.5, find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$.
Answer
Given:
$\text{P}(\text{A}\cap\text{B})=0.32$ and $\text{P(B)}=0.5$
We know that,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$=\frac{0.32}{0.5}$
$=\frac{16}{25}$
$=0.64$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.64$
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Question 1022 Marks
A bag contains 4 red and 5 black balls, a second bag contains 3 red and 7 black balls. One ball is drawn at random from each bag, find the probability that the,
Balls are of different colours.
Answer
Given,
Bag (1) contains 4 red and 6 black balls.
Bag (2) contains 3 red and 7 black balls
One ball is drawn ar random from each bag.
P (Balls are of different colours)
$=\text{P}\big((\text{R}_1\cap\text{B}_2)\cup(\text{B}_1\cap\text{R}_2)\big)$
$=\text{P}(\text{R}_1\cap\text{B}_2)+\text{P}(\text{B}_1\cap\text{R}_2)$
$=\text{P}(\text{R}_1)\text{P}(\text{B}_2)+\text{P}(\text{B}_1)\text{P}(\text{R}_2)$
$=\frac{4}{9}\times\frac{7}{10}+\frac{5}{9}\times\frac{9}{10}$
$=\frac{28}{90}+\frac{15}{90}$
$=\frac{43}{90}$
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Question 1032 Marks
If X follows binomial distribution with parameters n = 5, p and P(X = 2) = 9P(X = 3), then find the value of p.
Answer
We have,
X follows binomial distribution with parameters $\text{n = 5, p}$ and $\text{P(X}=2)=9\text{P(X}=3).$
So, $\text{P(X = r})=\text{ }^{5}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(5-\text{r})},$ where $\text{r}=0,1,2,3,4,5$ and $\text{q}=1-\text{p}$
As, $\text{P(X}=2)=9\text{P(X}=3)$
$\Rightarrow\text{ }^5\text{c}_2\text{p}^2\text{q}^3=9^5\text{c}_3\text{p}^3\text{q}^2$
$\Rightarrow10\text{p}^2\text{q}^3=9\times10\text{p}^3\text{q}^2$
$\Rightarrow\text{q}=9\text{p}$
$\Rightarrow1-\text{p}=9\text{p}$ [As, q = 1 - p]
$\Rightarrow10\text{p}=1$
$\therefore\text{p}=\frac{1}{10}$
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Question 1042 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
Answer
Given that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{0.18}{0.3}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.6$
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Question 1052 Marks
A fair die is rolled. Consider events E = $\{1,\ 3,\ 5\},\ \text{F}=\{2,\ 3\}\ \text{and}\ \text{G}=\{2,\ 3,\ 4,\ 5\}.\ \text{Find}:$
$\text{P}(\text{E}|\text{F})\ \text{and}\ \text{P}(\text{F}|\text{E})$
Answer
$\text{S}=(1,\ 2,\ 3,\ 4,\ 5,\ 6)\ \Rightarrow\ \ \ \ \ \ \text{n}(\text{S})=6$
$\text{E}=(1,\ 3,\ 5)\ \ \ \ \ \ \ \ \text{F}=(2,\ 3)\ \ \ \ \ \ \ \ (\text{G})=(2,\ 3,\ 4,\ 5)$
$\Rightarrow\ \ \ \ \ \text{n}(\text{E})=3\ \ \ \ \ \ \ \ \text{n}(\text{F})=2\ \ \ \ \ \ \ \text{n}(\text{G})=4$
$\text{P}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{3}{6}\ \ \ \ \ \ \ \ \ \text{P}\left(\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{2}{6}$
$\text{E}\cap\text{F}=\left(3\right)\ \Rightarrow\ \ \ \ \text{n}\left(\text{E}\cap\text{F}\right)=1$
$\text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\ \cap\ \text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{1}{6}$
$\text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac{1}{6}}{\frac{2}{6}}=\frac{1}{2}\ \ \ \ \ \ \ \ \\ \text{and}\ \ \ \text{P}\left(\text{F}|\text{E}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{F}\right)}{\text{P}\left(\text{E}\right)}=\frac{\frac{1}{6}}{\frac{3}{6}}=\frac{1}{3}$
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Question 1062 Marks
A bag contains 7 red, 5 white and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that
none is white?
Answer
Let P be the probability of getting 1 white ball out of 7 red, 5 white and 8 black balls. so
$\text{p}=\frac{5}{20}$
$\text{p}=\frac{1}{4}$
$\text{q}=1-\frac{1}{4}$ [Since p + q = 1]
Let X denote the random variable of number of selecting white ball with replacement out of 4 balls. probability of getting r white balls out of n balls is given by
$\text{P}(\text{x = r})=\text{ }^\text{n}\text{c}_\text{r}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^4\text{c}_{\text{r}}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{4-\text{r}}\dots(1)$
Probabitilty of getting none white ball
$=\text{P}(\text{x}=0)$
$=\text{ }^4\text{c}_0\big(\frac{1}{4}\big)^0\big(\frac{3}4{}\big)^{4-0}$ [Using (1)]
$=\big(\frac{3}{4}\big)^4$
$=\frac{81}{256}$
Probability of getting none white ball $=\frac{81}{256}$
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Question 1072 Marks
Find the expected number of boys in a family with 8 children, assuming the sex distribution to be equally probable.
Answer
Here, $\text{n}=8$
Let p be the probability of number of boys in the family.
$\text{p}=\frac{1}{2},\text{q}=\frac{1}{2}$
Expected number of boys = Mean
$\Rightarrow\text{np}=4$
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Question 1082 Marks
Which of the following distributions of a random variable X are the probability distributions?
X:
0
1
2
P(X):
0.6
0.4
0.2
Answer
P(X = 0) + P(X = 1) + P(X = 1) + P(X = 2)
= 0.6 + 0.4 + 0.2
= 1.2 > 1
It is not the probability distribution of random variable X.
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Question 1092 Marks
A card is drawn from a well-shulffled deck of 52 cards. The outcome is noted, the card is replaced and the deck reshuffled. Another card is then drawn from the deck.
What is the probability that the first card is an ace and the second card is a red queen?
Answer
P(first ace and second red queen) = P(ace card) × P(red queen)
$=\frac{4}{52}\times\frac{2}{52}$
$=\frac{1}{338}$
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Question 1102 Marks
Which of the following distributions of a random variable X are the probability distributions?
X: 0 1 2 3 4
P(X): 0.1 0.5 0.2 0.1 0.1
Answer
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= 0.1 + 0.5 + 0.2 + 0.1 + 0.1
= 1
It is the probability distribution of random variable X.
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Question 1112 Marks
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
E: ‘the card drawn is a spade’
F: ‘the card drawn is an ace’
Answer
In a deck of 52 cards, 13 cards are spades and 4 cards are aces.
$\therefore$ P(E) = P(the card drawn is a spade) $=\frac{13}{52}=\frac{1}{4}$
$\therefore$ P(F) = P(the card drawn is an ace)$=\frac{4}{52}=\frac{1}{13}$
In the deck of cards, only 1 card is an ace of spades.
P(EF) = P(the card drawn is spade and an ace) $=\frac{1}{52}$
$\text{P}(\text{E})\times\text{P}(\text{F})=\frac{1}{4}\cdot\frac{1}{13}=\frac{1}{52}=\text{P}(\text{EF})$
$\Rightarrow\text{P}(\text{E})\times\text{P}(\text{F})=\text{P}(\text{EF})$
Therefore, the events E and F are independent.
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Question 1122 Marks
Five cards are drawn successively with replacement from a well-shuffled pack of 52 cards. What is the probability that
none is a spade?
Answer
Let P denote the probability of geting one spade out of a deck of 52 cards, So
$\text{p}=\frac{13}{52}$
$\text{p}=\frac{1}{4}$
$\text{q}=1-\frac{1}{4}$ [Since p + q = 1]
$\text{q}=\frac{3}{4}$
let X denote the radom variable of number of spades out of 5 cards. probability of getting r spades out of n cards is given by
$\text{P}(\text{x = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^5\text{c}_\text{r}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{5-\text{r}}\dots(1)$
P(none is a spade) = P(X = 0)
$=\text{ }^5\text{C}_0\big(\frac{1}{4}\big)^0\big(\frac{3}{4}\big)^5$
$=\frac{243}{1024}$
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Question 1132 Marks
In a competition A, B and C are participating. The probability that A wins is twice that of B, the probability that B wins is twice that of C. Find the probability that A losses.
Answer
Let P(A wins) = x
So, P(B wins) = 2x
P(A wins) = 2P(B wins)
= 2(2x)
P(A wins) = 4X
P(A wins) + P(B wins) + P(C wins) = 1
⇒ 4x + 2x+ x = 1
⇒ 7x = 1
$\Rightarrow\ \text{x}=\frac{1}{7}$
$\text{P(A wins)}=4\text{x}$
$=\frac{4}{7}$
$\text{P(A losses)}=1-\text{P(A wins)}$
$=1-\frac{4}{7}$
$=\frac{3}{7}$
Required probability $=\frac{3}{7}$
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Question 1142 Marks
If X denotesthe number on the uper face of a cubical die when it is thrown, find the mean of X.
Answer
A cubical die can show 1, 2, 3, 4, 5 or 6 on its face
$\text{x}_\text{i}$
 $\text{p}_\text{i}$
 $\text{p}_\text{i}\text{x}_\text{i}$
$1$
 $\frac{1}{6}$
 $\frac{1}{6}$
$2$
 $\frac{1}{6}$
 $\frac{2}{6}$
$3$
 $\frac{1}{6}$
 $\frac{3}{6}$
$4$
 $\frac{1}{6}$
 $\frac{4}{6}$
$5$
 $\frac{1}{6}$
 $\frac{5}{6}$
$6$
 $\frac{1}{6}$
 $\frac{6}{6}$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{1}{6}+\frac{2}{6}+\frac{3}{6}+\frac{4}{6}+\frac{5}{6}+\frac{6}{6}=\frac{21}{6}=3.5$
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Question 1152 Marks
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.
Answer
Total number of balls = 18
Number of red balls = 8
Number of black balls = 10
Probability of getting first ball as red $=\frac{8}{18}=\frac{4}{9}$
The ball is replaced after the first draw.
Probability of getting second ball as black $=\frac{10}{18}=\frac{5}{9}$
Therefore, probability of getting first ball as black and second ball as red $=\frac{4}{9}\times\frac{5}{9}=\frac{20}{81}$
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
$=\frac{20}{81}+\frac{20}{81}$
$=\frac{40}{81}$
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Question 1162 Marks
A die is tossed thrice. Find the probability of getting an odd number at least once.
Answer
Probability of getting an odd number in a single throw of a die $=\frac{3}{6}=\frac{1}{2}$

Similarly, probability of getting an even number $\frac{3}{6}=\frac{1}{2}$

Probability of getting an even number three times $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}$

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

$=1-\frac{1}{8}$

$=\frac{7}{8}$

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Question 1172 Marks
The probability distribution of random variable X is given below:

$\text{X}$
$0$
$1$
$2$
$3$
$\text{P}(\text{X})$
$\text{k}$
$\frac{\text{k}}{2}$
$\frac{\text{k}}{4}$
$\frac{\text{k}}{8}$

 Determine $\text{P}(\text{X}\leq2)$ and $\text{P}(\text{X}>2)$

Answer
$\text{P}(\text{X}\leq2)$
$=\text{P}(0)+\text{P}(1)+\text{P}(2)$
$=\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}$
$=\frac{8}{15}+\frac{8}{30}+\frac{8}{60}$
$=\frac{14}{15}$
$\text{P}(\text{X}>2)=\text{P}(3)=\frac{\text{k}}{8}=\frac{1}{15}$
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