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

Questions · Page 3 of 3

2 Marks Questions

Question 1012 Marks

Kamal and Monica appeared for an interview for two vacancies. The probability of Kamal's selection is $\frac{1}{3}$ and that of Monika's selection is $\frac{1}{5}$. Find the probability that.
At least one of them will be selected.

Answer

P (Kamal gets selected) $=\text{P(A)}=\frac{1}{3}$
P (Monica gets selected) $=\text{P(B)}=\frac{1}{5}$
P (At least one of them gets selected) $=\text{P}(\text{A}\cup\text{B})$
$=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})$
$=\text{P(A)}+\text{P(B)}-\text{P(A)}\times\text{P(B)}$
$=\frac{1}{3}+\frac{1}{5}-\frac{1}{3}\times\frac{1}{5}$
$=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}$
$=\frac{7}{15}$

View full question & answer→

Question 1022 Marks

Three numbers are chosen from $1$ to $20$. Find the probability that they are consecutive.

Answer

$S =$ There numbers are chosen from $1$ to $20$
$n(S) = {^{20}C_3}$
$E =$ Group of three consecutive numbers between $1$ and $20$
$n(E) = 18$
$\{(1, 2, 3,), (2, 3, 4), (3, 4, 5), (4, 5, 6), (5, 6, 7), (6, 7, 8), (7, 8, 9), (8, 9, 10), (9, 10, 11), (10, 11, 12), (11, 12, 13), (11, 12, 13), (12, 13, 14), (13, 14, 15), (14, 15, 16), (15, 16, 17), (16, 17, 18), (17, 18, 19), (18, 19, 20)\}$
$\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}$
$=\frac{18}{^{20}\text{C}_3}$
Required probability $=\frac{18}{^{20}\text{C}_3}$

View full question & answer→

Question 1032 Marks

Two cards are drawn without replacement from a pack of 52 cards. Find the probability that.
The first is a king and the second is an ace.

Answer

We know that, there are 4 king and 4 ace in a pack of 52 cards.
Two cards are drawn without replacment
A = First catd is king
B = Second card an ace
P (The first card is a king second is an ace)
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{4}{52}\times\frac{4}{21}$
$=\frac{4}{663}$
Required probability $=\frac{4}{663}$

View full question & answer→

Question 1042 Marks

If $\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{11},$ find
$\text{P}(\text{A}\cap\text{B})$

Answer

Given,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11},\text{P}(\text{A}\cup\text{B})=\frac{7}{11}$
Since, $\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})$
$=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}$
$\text{P}(\text{A}\cap\text{B})=\frac{4}{11}$

View full question & answer→

Question 1052 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
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P(A) }\text{P(B)}$
$=\big[1-\text{P(A)}\big]\big[1-\text{P(B)}\big]$
$=(1-0.3)(1-0.6)$
$=0.7\times0.4$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=0.28$

View full question & answer→

Question 1062 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

$\text{S}=(1,\ 2,\ 3,\ 4,\ 5,\ 6)\ \Rightarrow\ \ \ \ \ \ \text{n}(\text{S})=6$
$\text{E}=(1,\ 3,\ 5)\ \ \ \ \ \ \ \ \text{F}=(2,\ 3)\ \ \ \ \ \ \ \ (\text{G})=(2,\ 3,\ 4,\ 5)$
$\Rightarrow\ \ \ \ \ \text{n}(\text{E})=3\ \ \ \ \ \ \ \ \text{n}(\text{F})=2\ \ \ \ \ \ \ \text{n}(\text{G})=4$
$\text{P}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{3}{6}\ \ \ \ \ \ \ \ \ \text{P}\left(\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{2}{6}$
$\text{E}\cap\text{F}=\left(3\right)\ \Rightarrow\ \ \ \ \text{n}\left(\text{E}\cap\text{F}\right)=1$
$\text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\ \cap\ \text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{1}{6}$
$\text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac{1}{6}}{\frac{2}{6}}=\frac{1}{2}\ \ \ \ \ \ \ \ \\ \text{and}\ \ \ \text{P}\left(\text{F}|\text{E}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{F}\right)}{\text{P}\left(\text{E}\right)}=\frac{\frac{1}{6}}{\frac{3}{6}}=\frac{1}{3}$

View full question & answer→

Question 1072 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 $=\frac{81}{256}$

View full question & answer→

Question 1082 Marks

In a group of 200 items, if the probability of getting a defective item is 0.2, write the mean of the distribution.

Answer

Let p be the probability of defective item, so
$\text{p}=0.2$
$\text{p}=\frac{1}{5}$
$\text{q}=1-\frac{1}{5}$ [Since p+ q =1]
$\text{q}=\frac{4}{5}$
Given, $\text{n}=200$
$\text{Mean = np}$
$=200\big(\frac{1}{5}\big)$
$=40$
$\text{Mean}=40$

View full question & answer→

Question 1092 Marks

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that.
Both the balls are red.

Answer

Given,
Box contains 10 black and 8 red balls.
Two balls are drawn with replacement.
P (Both the balls are red)
$=\text{P}(\text{R}_1\cap\text{R}_2)$
$=\text{P}(\text{R}_1)\text{ P}(\text{R}_2)$
$=\frac{8}{18}\times\frac{8}{18}$
$=\frac{16}{81}$
Required probability $=\frac{16}{81}$

View full question & answer→

Question 1102 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.

View full question & answer→

Question 1112 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)
$=\frac{4}{52}\times\frac{2}{52}$
$=\frac{1}{338}$

View full question & answer→

Question 1122 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 $=\sum\text{p}_\text{i}\text{x}_\text{i}=0.5\times\text{k}+1\times\text{k}^2+1.5\times2\text{k}^2+2\times\text{k}$
$=0.5\times\frac{1}{3}+1\times\Big(\frac{1}{3}\Big)^2+1.5\times2\Big(\frac{1}{3}\Big)^2+2\times\frac{1}{3}$
$=\frac{0.5}{3}+\frac{1}{9}+\frac{3}{9}+\frac{2}{3}$
$=\frac{1.5+1+3+6}{9}$
$=\frac{11.5}{9}$
$=\frac{115}{90}$
$=\frac{23}{18}$

View full question & answer→

Question 1132 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)
$=\text{ }^5\text{C}_0\big(\frac{1}{4}\big)^0\big(\frac{3}{4}\big)^5$
$=\frac{243}{1024}$

View full question & answer→

Question 1142 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,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$=\frac{0.32}{0.5}$
$=\frac{16}{25}$
$=0.64$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.64$

View full question & answer→

Question 1152 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)
$=\text{P}\big((\text{R}_1\cap\text{B}_2)\cup(\text{B}_1\cap\text{R}_2)\big)$
$=\text{P}(\text{R}_1\cap\text{B}_2)+\text{P}(\text{B}_1\cap\text{R}_2)$
$=\text{P}(\text{R}_1)\text{P}(\text{B}_2)+\text{P}(\text{B}_1)\text{P}(\text{R}_2)$
$=\frac{4}{9}\times\frac{7}{10}+\frac{5}{9}\times\frac{9}{10}$
$=\frac{28}{90}+\frac{15}{90}$
$=\frac{43}{90}$

View full question & answer→

Question 1162 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]
$\Rightarrow10\text{p}=1$
$\therefore\text{p}=\frac{1}{10}$

View full question & answer→

Question 1172 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
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{0.18}{0.3}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.6$

View full question & answer→

Generate Paper Free All Probability topics

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