2 Marks Questions - Page 3 - MATHS STD 12 Science Questions - Vidyadip

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2 Marks Questions

Question 1012 Marks

A discrete random variable $X$ has the probability distribution given below:

$X:$	$0.5$	$1$	$1.5$	$2$
$P(X):$	$k$	$k^2$	$2k^2$	$k$

Determine the mean of the distribution.

Answer

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=0.5\times\text{k}+1\times\text{k}^2+1.5\times2\text{k}^2+2\times\text{k}$
$=0.5\times\frac{1}{3}+1\times\Big(\frac{1}{3}\Big)^2+1.5\times2\Big(\frac{1}{3}\Big)^2+2\times\frac{1}{3}$
$=\frac{0.5}{3}+\frac{1}{9}+\frac{3}{9}+\frac{2}{3}$
$=\frac{1.5+1+3+6}{9}$
$=\frac{11.5}{9}$
$=\frac{115}{90}$
$=\frac{23}{18}$

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Question 1022 Marks

Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$

Answer

Given that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P(A) }\text{P(B)}$
$=\big[1-\text{P(A)}\big]\big[1-\text{P(B)}\big]$
$=(1-0.3)(1-0.6)$
$=0.7\times0.4$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=0.28$

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Question 1032 Marks

Fatima and John appear in an interview for two vacancies for the same post. The probability of Fatima's selection is $\frac{1}{7}$ and that of John's selection is $\frac{1}{5}$. What is the probability that,
None of them will be selected?

Answer

Given,
Probability of Fatima's (F) selection $=\frac{1}{7}$
$\text{P(F)}=\frac{1}{7}\Rightarrow\ \text{P}(\overline{\text{F}})=\frac{6}{7}$
Probability of John's (J) selection $=\frac{1}{5}$
$\text{P(F)}=\frac{1}{5}\Rightarrow\ \text{P}(\overline{\text{F}})=\frac{4}{5}$
P(None of them selected)
$=\text{P}(\overline{\text{F}}\cap\overline{\text{J}})$
$=\text{P}(\overline{\text{F}})+\text{P}(\overline{\text{J}})$
$=\frac{6}{7}\times\frac{4}{5}$
$=\frac{24}{35}$
Required probabilty $=\frac{24}{35}$

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Question 1042 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 1052 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 1062 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 1072 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 1082 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 1092 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 1102 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 1112 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 1122 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 1132 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 1142 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 1152 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 1162 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 1172 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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2 Marks Questions - Page 3 - MATHS STD 12 Science Questions - Vidyadip