Probability(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

PROBABILITY MCQs

Probability, Probability I

Total Questions : 775 | Page 13 of 78 pages

Question 121. 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

3255
2155
1955
None of these

Discuss Question

Answer: Option D. -> None of these
:
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 122. Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals

Discuss Question

Answer: Option B. -> 715
:
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 123. 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

Discuss Question

Answer: Option B. -> 16
:
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 124. 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?

Discuss Question

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 125. 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

Discuss Question

Answer: Option D. -> 3:2
:
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 126. 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

Discuss Question

Answer: Option B. -> 218
:
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}

\frac{2}{18}

Question 127. 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

Discuss Question

Answer: Option B. -> 817
:
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}

3^{r d}

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

Discuss Question

Answer: Option C. -> 37
:
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}

Question 129. If A and B are two events such that P(A) =

\frac{1}{2}

and P(B) =

\frac{2}{3}

, then

P(A∪B)≥23
16≤P(A′∩B)≤12
16≤P(A∩B)≤12
All of the above

Discuss Question

Answer: Option D. -> All of the above
:
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 130. 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