Probability I(11th And 12th > Mathematics ) Questions and answers for exam Preparation

Question 1.

The probability that A speaks truth is $\frac{4}{5}$ , while this probability for B is $\frac{3}{4}$ . The probability that they contradict each other when asked to speak on a fact is

$\frac{3}{23}$
$\frac{1}{20}$
$\frac{1}{5}$
$\frac{14}{40}$

Explanation:-

Answer: Option D. ->

\frac{14}{40}

:
D

In short the event described here is P(A)'P(B) +P(A)P(B)'

where P(A) and P(B) are the probabilities of A and B speaking the truth respectively.

P(E)= P(A)'P(B) +P(A)P(B)' = $\frac{1}{5} \times \frac{3}{4} + \frac{4}{5} \times \frac{1}{4}$

P(E)= $\frac{3}{20} + \frac{4}{20} = \frac{7}{20} = \frac{14}{40}$

Question 2.

Ashu studies at Byju's classes and her probability of selection in IIT-JEE is $\frac{4}{5}$ . Ridhima took coaching at FIIT-JEE and the probability of her selection is $\frac{2}{3}$ . What is the probability that only 1 of them cracks the Exam?

$\frac{2}{5}$
$\frac{3}{10}$
$\frac{4}{15}$
None of the above.

Explanation:-

Answer: Option A. ->

\frac{2}{5}

:
A

P(A)= $\frac{4}{5}$

P(R)= $\frac{2}{3}$

P(E)= P(A'R)+P(AR')

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

P(E)= $\frac{2}{15} + \frac{4}{15}$

P(E)= $\frac{6}{15}$

P(E)= $\frac{2}{5}$

Question 3.

Probability of a fraudster being caught is $\frac{1}{2}$ and committing a fraud is $\frac{3}{5}$ . What is his chance of not going to jail?

70%
50%
40%
35%

Explanation:-

Answer: Option A. -> 70%
:
A

P(not going to jail)= P( F')+ P(F intersection C')

where P(F) is the probability of commiting a fraud and P(C) is the Probability of being caught

P(F')= $\frac{2}{5}$

P(C')= $\frac{1}{2}$

P(F intersection C')= P(F) × P(C')= $\frac{3}{5} \times \frac{1}{2} = \frac{3}{10}$

P(not going to jail)= $\frac{2}{5} + \frac{3}{10} = \frac{7}{10} = 70 %$

Question 4.

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is

$\frac{2}{9}$
$\frac{2}{18}$
$\frac{3}{8}$
$\frac{1}{20}$

Explanation:-

Answer: Option B. ->

\frac{2}{18}

:
B

For a particular house being selected Probability = $\frac{1}{3}$

Prob(all the persons apply for the same house) = $(\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}) \times 3 = \frac{1}{9}$ or $\frac{2}{18}$

Question 5.

Two persons A and B take turns in throwing a pair of dice. The first person to through 9 from both dice will be awarded the prize. If A throws first then the probability that B wins the game is

$\frac{9}{17}$
$\frac{8}{17}$
$\frac{8}{9}$
$\frac{1}{9}$

Explanation:-

Answer: Option B. ->

\frac{8}{17}

:
B
The probability of throwing 9 with two dice =

\frac{4}{36} = \frac{1}{9}

∴

The probability of not throwing 9 with two dice =

\frac{8}{9}

If A is to win he should throw 9 in

1^{s t}

or

3^{r d}

or

5^{t h}

attempt
If B is to win, he should throw, 9 in

2^{n d}

,

4^{t h}

attempt
B’s chances =

(\frac{8}{9}) . \frac{1}{9} + {(\frac{8}{9})}^{3} . \frac{1}{9} + . . . . . = \frac{\frac{8}{9} \times \frac{1}{9}}{1 - {(\frac{8}{9})}^{2}} = \frac{8}{17}

Question 6.

There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, is a random order till both the faulty machines are identified. Then the probability that only two tests are needed

$\frac{1}{3}$
$\frac{1}{6}$
$\frac{1}{2}$
$\frac{1}{4}$

Explanation:-

Answer: Option B. ->

\frac{1}{6}

:
B
This is a problem of without replacement.

P = \frac{o n e d e f . f r o m 2 d e f .}{a n y o n e f r o m 4} \times \frac{1 d e f . f r o m r e m a i n i n g 1 d e f .}{a n y o n e f r o m r e m a i n i n g 3}

Hence required probability =

\frac{2}{4} \times \frac{1}{3} = \frac{1}{6}

Aliter : Number of ways in which two faulty machines may be detected (depending upon the test done to identify the faulty machines) =

^{4} C_{2} = 6

Number of favourable cases = 1
[When faulty machines are identified in the first and the second test].
Hence required probability =

\frac{1}{6}

Question 7.

In four schools $B_{1}, B_{2}, B_{3}, B_{4}$ the percentage of girls students is 12, 20, 13, 17 respectively. From a school selected at random, one student is picked up at random and it is found that the student is a girl. The probability that the school selected is $B_{2}$ , is

$\frac{6}{31}$
$\frac{10}{31}$
$\frac{13}{62}$
$\frac{17}{62}$

Explanation:-

Answer: Option B. ->

\frac{10}{31}

:
B
Favorable number of cases =

^{20} C_{1} = 20

Sample space =

^{62} C_{1} = 62

∴

Required probability =

\frac{20}{62} = \frac{10}{31}

Question 8.

Cards are drawn one by one at random from a well shuffled full pack of 52 cards until two aces are obtained for the first time. If N is the number of cards required to be drawn, then $P_{r} N = n$ where $2 \leq n \leq 50,$ is

$\frac{(n - 1) (52 - n) (51 - n)}{50 \times 49 \times 17 \times 3}$
$\frac{2 (n - 1) (52 - n) (51 - n)}{50 \times 49 \times 17 \times 3}$
$\frac{3 (n - 1) (52 - n) (51 - n)}{50 \times 49 \times 17 \times 3}$
$\frac{4 (n - 1) (52 - n) (51 - n)}{50 \times 49 \times 17 \times 3}$

Explanation:-

Answer: Option A. ->

\frac{(n - 1) (52 - n) (51 - n)}{50 \times 49 \times 17 \times 3}

:
A
Here the least number of draws to obtain 2 aces are 2 and the maximum number is 50 thus n can take value from 2 to 50.
Since we have to make n draws for getting two aces, in (n – 1) draws, we get any one of the 4 aces and in the

n^{t h}

draw we get one ace. Hence the required probability

= \frac{^{4} C_{1} \times^{48} C_{n - 2}}{^{52} C_{n - 1}} \times \frac{3}{52 - (n - 1)}

= \frac{4 \times (48)!}{(n - 2)! (48 - n + 2)!} \times \frac{(n - 1)! (52 - n + 1)!}{(52)!} \times \frac{3}{52 - n + 1}

= \frac{(n - 1) (52 - n) (51 - n)}{50 \times 49 \times 17 \times 3}

(on simplification).

Question 9.

Six boys and six girls sit in a row. What is the probability that the boys and girls sit alternatively

$\frac{1}{462}$
$\frac{1}{924}$
$\frac{1}{2}$
None of these

Explanation:-

Answer: Option A. ->

\frac{1}{462}

:
A
Let n = total number of ways = 12!
and m = favourable numbers of ways =

2 \times 6! . 6!

Since the boys and girls can sit alternately in 6! . 6! ways if we begin with a boy and similarly
they can sit alternately in 6! . 6! Ways if we begin with a girl
Hence required probability =

\frac{m}{n} = \frac{2 \times 6! . 6!}{12!} = \frac{1}{462}

Question 10.

Let X be a set containing n elements. If two subsets A and B of X are picked at random, the probability that A and B the same number of elements, is

$\frac{^{2 n} C_{n}}{2^{2 n}}$
$\frac{1}{^{2 n} C_{n}}$
$\frac{1.3.5.... (2 n - 1)}{2^{n}}$
$\frac{3^{n}}{4^{n}}$

Explanation:-

Answer: Option A. ->

\frac{^{2 n} C_{n}}{2^{2 n}}

:
A
We know that the number of sub-sets of a set containing n elements is

2^{n}

.
Therefore the number of ways of choosing A and B is

2^{n} . 2^{n} = 2^{2 n}

We also know that the number of sub-sets (of X) which contain exactly r elements is

^{n} C_{r} .

Therefore the number of ways of choosing A and B, so that they have the same number elements is

(^{n} C_{0})^{2} + (^{n} C_{1})^{2} + (^{n} C_{2})^{2} + . . . . + (^{n} C_{n})^{2} =^{2 n} C_{n}

Thus the required probability =

\frac{^{2 n} C_{n}}{2^{2 n}} .