Free Objective Test 02 Practice Test - 11th and 12th

Question 1

A random variable X has the following probability distribution
$\begin{matrix} X & P (X = x) & X & P (X = x) 0 & λ & 5 & 11 λ 1 & 3 λ & 6 & 13 λ 2 & 5 λ & 7 & 15 λ 3 & 7 λ & 8 & 17 λ 4 & 9 λ \end{matrix}$
then, $λ$ is equal to

\frac{1}{81}

\frac{2}{81}

\frac{5}{81}

\frac{7}{81}

SOLUTION

Solution : A

$\sum_{x = 0}^{8} P (X) = 1 \Rightarrow P (0) + P (1) + P (2) + P (3) + P (4) + P (5) + P (6) + P (7) + P (8) = 1 = λ + 3 λ + 5 λ + 7 λ + 9 λ + 11 λ + 13 λ + 15 λ + 17 λ = 1 \Rightarrow \frac{9}{2} (λ + 17 λ) = 1 \Rightarrow 81 λ = 1 ∴ λ = \frac{1}{81}$

Question 2

A man is known to speak truth is 75% cases. If he throws an unbiased die and tells his friend that it is a six, then the probability that it is actually a six, is

\frac{1}{6}

\frac{1}{8}

\frac{3}{4}

\frac{3}{8}

SOLUTION

Solution : D

Let E denote the event that a six occurs and A be the event theat the man reports that it is a six. We have $P (E) = \frac{1}{6}, P (E^{'}) = \frac{5}{6}, P (\frac{A}{E}) = \frac{3}{4}$ and $P (\frac{A}{E^{'}}) = \frac{1}{4}$ . By Baye’s theorem
$P (\frac{E}{A}) = \frac{P (E) . P (\frac{A}{E})}{P (E) . P (\frac{A}{E}) + P (E^{'}) . P (\frac{A}{E^{'}})} = \frac{(\frac{1}{6}) (\frac{3}{4})}{(\frac{1}{6}) (\frac{3}{4}) + (\frac{5}{6}) (\frac{1}{4})} = \frac{3}{8}$

Question 3

Seven coupons are selected at random one at a time with replacement from 15 coupons numbered 1 to 15. The probability that the largest number appearing on a selected coupon is 9, is

{(\frac{9}{16})}^{6}

{(\frac{8}{15})}^{7}

{(\frac{3}{5})}^{7}

{(\frac{3}{5})}^{7}

{(\frac{8}{15})}^{7}

SOLUTION

Solution : D

Each coupon can be selected in 15 ways. The total number of ways of choosing 7 copouns is $15^{7}$ . If largest number is 9, then the selected numbers have to be from 1 to 9 excluding those consisting of only 1 to 8.
Probability desired is $\frac{9^{7} - 8^{7}}{15^{7}}$
$= {(\frac{3}{5})}^{7} - {(\frac{8}{15})}^{7}$

Question 4

Team A plays with 5 other teams exactly once. Assuming that for each match the probabilities of a win, draw and loss are equal, then

A. The probability that A wins and loses equal number of matches is

\frac{34}{81}

B. The probability that A wins and loses equal number of matches is

\frac{17}{81}

C. The probability that A wins more number of matches than it loses is

\frac{17}{81}

D. The probability that A loses more number of matches than it wins is

\frac{16}{81}

SOLUTION

Solution : B

Probability of equal number of W and L is the probability of following
(0)W,(0)L +(1)W,(1)L+(2)W, (2)L
$= {(\frac{1}{3})}^{5} +^{5} C_{1} .^{4} C_{1} {(\frac{1}{3})}^{5} +^{5} C_{2} .^{3} C_{2} {(\frac{1}{3})}^{5} = \frac{17}{81}$

Question 5

The probabilities of three mutually exclusive events A, B, C are given by $\frac{2}{3}, \frac{1}{4}, a n d \frac{1}{6}$ respectively. This statement

A. Is true

B. is false

C. nothing can be said

D. could be either

SOLUTION

Solution : B

Since the events A,B,C are mutually exclusive, we have
$P (A ⋃ B ⋃ C) = \frac{2}{3} + \frac{1}{4} + \frac{1}{6} = \frac{13}{12} > 0$
Which is not possible , hence the statement is false.

Question 6

Two events A and B have the probabilities 0.25 and 0.5 respectively. The probability that both A and B simultaneously is 0.12. then the probability that neither A and nor B occurs is ___.

A. 0.13

B. 0.38

C. 0.63

D. 0.37

SOLUTION

Solution : D

Given: $P (A) = 0.25$
$P (B) = 0.5$
P(A and B simultaneously happening) = 0.12

$P (A \cup B) = P (A) + P (B) - P (A \cap B) P (A \cup B) = 0.25 + 0.5 - 0.12 = 0.63$
$P (¯ A \cap ¯ B) = P ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ (A \cup B) = 1 - P (A \cup B) = 1 - 0.63 = 0.37$

Question 7

If $\frac{(1 + 4 p)}{4}, \frac{(1 - p)}{2}, and \frac{(1 - 2 p)}{2}$ are the probabilities of three mutually exclusive events , then the value of p is

\frac{1}{2}

\frac{1}{3}

\frac{1}{4}

\frac{1}{5}

SOLUTION

Solution : A

$\frac{1 + 4 p}{4}, \frac{1 - p}{2}, \frac{1 - 2 p}{2}$ are the probabilities of three mutually exclusive events.

$∴ 0 \leq \frac{1 + 4 p}{p} \leq 1, 0 \leq \frac{1 - p}{2} \leq 1$

$0 \leq \frac{1 - 2 p}{2} \leq 1 a n d 0 \leq (\frac{1 + 4 p}{4}) + (\frac{1 - p}{2}) + \frac{1 - 2 p}{2} \leq 1$

$∴ \frac{- 1}{4} \leq p \leq \frac{3}{4} 1 - \leq p \leq 1, \frac{1}{2} \leq p \leq \frac{1}{2} a n d p \leq \frac{5}{2}$

$\Rightarrow p = \frac{1}{2}$

Question 8

A father has three children with at least one boy. The probability that he has two boys and one girl is

\frac{1}{4}

\frac{3}{7}

\frac{1}{3}

\frac{2}{5}

SOLUTION

Solution : B

Let A be the event that father has at least one boy and B be the event that he has 2boys and one girl.
$P (A) = P (1 b o y, 2 g i r l) + P (2 b o y, 1 g i r l) + P (3 b o y, n o g i r l) = \frac{3 C_{1} + 3 C_{2} + 3 C_{3}}{2^{3}} = \frac{7}{8}$
$P (A ⋂ B) = P (2 b o y, 1 g i r l) = \frac{3}{8}$
$H e n c e P (\frac{B}{A}) = \frac{P (A ⋂ B)}{P (A)} = \frac{3}{7} .$

Question 9

A fair coin is tossed n times and let X denotes the number of heads obtained. If P(X = 4), P(X = 5), and P(X = 6) are in A.P., then n is equal to

A. Only 7

B. 14

C. 7 or 14

D. 8

SOLUTION

Solution : B and C

Probability of r successes in n trials is equal to $^{n} C_{r} p^{r} q^{n - r}$ , where p is the probability of success in one trial and q is the probability of failure in one trial.
This means $n C_{4} {(\frac{1}{2})}^{n}, n C_{5} {(\frac{1}{2})}^{n}, n C_{6} {(\frac{1}{2})}^{n} a r e A . P .$
$\Rightarrow 2 \times n C_{5} = n C_{4} + n C_{6}$
$\Rightarrow 2 \times \frac{n!}{(n - 5)! 5!} = \frac{n!}{(n - 4)! 4!} + \frac{n!}{(n - 6)!}!$
$\Rightarrow 6 \times 5 + (n - 4) (n - 5) = 2 \times 6 (n - 4)$
$\Rightarrow n^{2} - 9 n + 50 = 2 (6 n - 24)$
$\Rightarrow n^{2} - 21 n + 98 = 0$
$\Rightarrow n = 7 o r n = 14.$

Question 10

One hundred identical coins, each with probability p of showing up heads are tossed once. If 0 < p < 1 and the probability of heads showing 50 coins is equal to that head showing 51 coins, then the value of p is

\frac{1}{2}

\frac{49}{101}

\frac{50}{101}

\frac{51}{101}

SOLUTION

Solution : D

Let P1 and P2 be the respectively be the probability of heads showing 50 coins and head showing 51 coins $P_{1} = 100 C_{50} p^{50} (1 - p)^{50} a n d P_{2} = 100 C_{51} p^{51} (1 - p)^{49}$
$G i v e n P_{1} = P_{2}$
$\Rightarrow 100 C_{50} p^{50} (1 - p)^{50} = 100 C_{51} p^{51} (1 - p)^{49}$
$\Rightarrow \frac{1 - p}{p} = \frac{50}{51} \Rightarrow p = \frac{51}{101} .$

Question 11

Three six faced fair dice are rolled together. The probability that the sum of the numbers appearing on the dice is 8, is

\frac{7}{72}

\frac{7}{54}

\frac{4}{27}

\frac{7}{64}

SOLUTION

Solution : A

Getting sum 8 when three dice are rolled = E
The possibilities are (1,1,6) , (2,2,4), (3,3,2), (1,2,5), (1,3,4)
The number of favorable cases for E = 3 + 3 + 3 + 6 + 6 = 21
The total number of cases = $6^{3}$ = 216
Required probability P(E) $= \frac{21}{216} = \frac{7}{72} .$

Question 12

In a multiple choice question there are four alternative answers of which one or more than one is or are correct. A candidate will get marks on the question only if he ticks all correct answers. The candidate decides to tick answers at random. If he is allowed up to three chances to answer the question, the probability that he will get marks on it is given by

1 - {(\frac{14}{15})}^{3}

{(\frac{1}{15})}^{3}

\frac{1}{5}

\frac{14}{15}

SOLUTION

Solution : C

The total number of ways of answering one or more alternatives of 4 is $4 C_{1} + 4 C_{2} + 4 C_{3} + 4 C_{4} = 15,$
Out of these 15 combinations, only one combination is correct. The probability of answering the alternaive correctly at the first trial is $\frac{1}{15},$ that at the second trial is $(\frac{14}{15} . \frac{1}{14}) = \frac{1}{15,}$ and that at the third trail is $\frac{14}{15} . \frac{13}{14} . \frac{1}{13} = \frac{1}{15},$
Therefore the probability that the andidate will get marks on the question if he allowed upto three chances, is $\frac{1}{15} + \frac{1}{15} + \frac{1}{15} = \frac{1}{5} .$

Question 13

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

\frac{32}{55}

\frac{30}{40}

\frac{32}{50}

\frac{3}{4}

SOLUTION

Solution : A

case(i): red ball from A to B, red ball from B to A ,then red ball from A
$P_{1} = \frac{6}{10} \times \frac{5}{11} \times \frac{6}{10}$
Case(ii): red ball from A to B, white ball from B to A, red ball from A
$P_{2} = \frac{6}{10} \times \frac{6}{11} \times \frac{5}{10}$
Case(iii): white ball from A to B, red ball from B to A, red ball from A
$P_{3} = \frac{4}{10} \times \frac{4}{11} \times \frac{7}{10}$
Case(iv): white ball from A to B, white ball from B to A, red ball from A
$P_{4} = \frac{4}{10} \times \frac{7}{11} \times \frac{6}{10}$
Therefore required probability $= P_{1} + P_{2} + P_{3} + P_{4} = \frac{640}{1100} = \frac{32}{55}$

Question 14

From a bag containing 9 distinct white and 9 distinct black, 9 balls are drawn at random one by one, the drawn balls being replaced each time. The probability that at least four balls of each colour is in the draw, is

A. A little less than

\frac{1}{2}

B. a little greater than

\frac{1}{2}

\frac{1}{2}

\frac{1}{3}

SOLUTION

Solution : A

Out of 9 distinct black and 9 distinct white balls, probability of drawing a white ball $= \frac{1}{2}$
and drawing black ball is also $\frac{1}{2} .$ for at least 4 of each colour in 9 draws with replacement, there are two cases,
Case(i)
P(getting 5 white, 4 black) $= 9 C_{5} {(\frac{1}{2})}^{9}$
Cases (ii)
P(getting 4 white, 5 black) $= 9 C_{4} {(\frac{1}{2})}^{9}$
These are exclusive so
P(atleast 4 of each colour) $= 9 C_{5} {(\frac{1}{2})}^{9} + 9 C_{4} {(\frac{1}{2})}^{9} = \frac{63}{128}$ which is little less than $\frac{1}{2}$

Question 15

Let A, B , and C be three independent events with $P (A) = \frac{1}{3}, P (B) = \frac{1}{2}, a n d P (C) = \frac{1}{4} .$ . The probability of exactly 2 of these events occurring, is equal to

\frac{1}{4}

\frac{7}{24}

\frac{3}{4}

\frac{17}{24}

SOLUTION

Solution : A

P(exactly two of A, B and C) = P(AB) + P(BC) + P(CA) – 3 P(ABC)
Since A, B ,C are independent events,
Required probability = P(A)P(B) + P(B)P(C) +P(C)P(A) – 3 P(A)P(B)P(C)
$= \frac{1}{3} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{4} + \frac{1}{4} \times \frac{1}{3} - 3 \frac{1}{3} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{4.}$

Free Objective Test 02 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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