Free Objective Test 01 Practice Test - 11th and 12th

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

$\frac{3}{23}$

$\frac{1}{20}$

$\frac{1}{5}$

$\frac{14}{40}$

SOLUTION

Solution : 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

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}

SOLUTION

Solution : 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 3

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.

SOLUTION

Solution : 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 4

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?

A. 70%

B. 50%

C. 40%

D. 35%

SOLUTION

Solution : 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 5

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}

SOLUTION

Solution : 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 6

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}

SOLUTION

Solution : 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 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}

SOLUTION

Solution : 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

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}

D. None of these

SOLUTION

Solution : 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 9

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}

SOLUTION

Solution : 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 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}}

SOLUTION

Solution : 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 = $\frac{^{2 n} C_{n}}{2^{2 n}} .$

Question 11

Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals

\frac{1}{2}

\frac{7}{15}

\frac{2}{15}

\frac{1}{3}

SOLUTION

Solution : B

The number of ways to arrange 7 white an 3 black balls in a row
$\frac{10!}{7! .3!} = \frac{10.9.8}{1.2.3} = 120$
Numbers of blank places between 7 balls are 6. There is 1 place before first ball and 1 place after last ball. Hence total number of places are 8.
Hence 3 black balls are arranged on these 8 places so that no two black balls are together in number of ways.
$=^{8} C_{3} = \frac{8 \times 7 \times 6}{1 \times 2 \times 3} = 56$
So required probability = $\frac{56}{120} = \frac{7}{15}$

Question 12

One of the two events must occur. If the chance of one is $\frac{2}{3}$ of the other, then odds in favour of the other are

2 : 3

1 : 3

3 : 1

3 : 2

SOLUTION

Solution : D

Let p be the probability of the other event, then the probability of the first event is $\frac{2}{3}$ p. Since two events are totally exclusive, we have $p + (\frac{2}{3}) p = 1 \Rightarrow p = \frac{3}{5}$
Hence odds in favour of the other are 3 : 5 – 3, i.e, 3 : 2.

Question 13

Urn A contains 6 red and 4 black balls and urn B contains 4 red and 6 black balls. One ball is drawn at random from urn A and placed in urn B. Then one ball is drawn at random from urn B and placed in urn A. If one ball is now drawn at random from urn A, the probability that it is found to be red, is

\frac{32}{55}

\frac{21}{55}

\frac{19}{55}

D. None of these

SOLUTION

Solution : D

Let the events are
$R_{1}$ = A red ball is drawn from urn A and placed in B
$B_{1}$ = A black ball is drawn from urn A and placed in B
$R_{2}$ = A red ball is drawn from urn A and placed in A
$B_{2}$ = A black ball is drawn from urn A and placed in B
R = A red ball is drawn in the second attempt from A
Then the required probability
$= P (R_{1} R_{2} R) + (R_{1} B_{2} R) + P (B_{1} R_{2} R) + P (B_{1} B_{2} R)$
$= P (R_{1}) P (R_{2}) P (R) + P (R_{1}) P (B_{2}) P (R) + P (B_{1}) P (R_{2}) P (R) + P (B_{1}) P (B_{2}) P (R)$
$= \frac{6}{10} \times \frac{5}{11} \times \frac{6}{10} \times \frac{6}{10} \times \frac{6}{11} \times \frac{5}{10} \times \frac{4}{10} \times \frac{4}{11} \times \frac{7}{10} + \frac{4}{10} \times \frac{7}{11} \times \frac{6}{10}$
$= \frac{32}{55}$

Question 14

If A and B are two events such that P(A) = $\frac{1}{2}$ and P(B) = $\frac{2}{3}$ , then

P (A \cup B) \geq \frac{2}{3}

\frac{1}{6} \leq P (A^{'} \cap B) \leq \frac{1}{2}

\frac{1}{6} \leq P (A \cap B) \leq \frac{1}{2}

D. All of the above

SOLUTION

Solution : D

We have $P (A \cup B) \geq$ max. ${P (A), P (B)} = \frac{2}{3}$
$P (A \cap B) \leq$ min. ${P (A), P (B)} = \frac{1}{2}$
and $P (A \cap B) = P (A) + P (B) - P (A \cup B) \geq P (A) + P (B) - 1 = \frac{1}{6}$
$\Rightarrow \frac{1}{6} \leq P (A \cap B) \leq \frac{1}{2}$
$P (A^{'} \cap B) = P (B) - P (A \cap B)$
Hence $\frac{2}{3} - \frac{1}{2} \leq P (A^{'} \cap B) \leq \frac{2}{3} - \frac{1}{6}$
$\Rightarrow \frac{1}{6} \leq P (A^{'} \cap B) \leq \frac{1}{2}$

Question 15

The probability that a leap year selected at random contains either 53 Sundays or 53 Mondays is ____.

\frac{2}{7}

\frac{4}{7}

\frac{3}{7}

\frac{1}{7}

SOLUTION

Solution : C

A leap year consists of 366 days comprising of 52 weeks and 2 days. There are 7 possibilities for these 2 extra days.
(i) Sunday, Monday
(ii) Monday, Tuesday
(iii) Tuesday, Wednesday
(iv) Wednesday, Thursday
(v) Thursday, Friday
(vi) Friday, Saturday
(vii) Saturday, Sunday
Let us consider two events :
A : the leap year contains 53 Sundays
B : the leap year contains 53 Mondays.
Then we have $P (A) = \frac{2}{7}, P (B) = \frac{2}{7}, P (A \cap B) = \frac{1}{7}$
$∴$ Required probability = $P (A \cup B)$
$= P (A) + P (B) - P (A \cup B) = \frac{2}{7} + \frac{2}{7} - \frac{1}{7} = \frac{3}{7}$

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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