Probability(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

PROBABILITY MCQs

Probability, Probability I

Total Questions : 775 | Page 15 of 78 pages

Question 141. On a toss of two dice, A throws a total of 5. Then the probability that he will throw another 5 before he throws 7 is

Discuss Question

Answer: Option B. -> 25
:
B
let P(A) be the probability of throwing total of 5 and P(B) the probability of throwing total 7.

∴ P (A) = \frac{4}{36} = \frac{1}{9}, P (B) = \frac{6}{36} = \frac{1}{6}, P (A o r B) = \frac{5}{18} a n d P (A a n d B) = 0 P (n e i t h e r A n o r B) = \frac{13}{18}, ∴ P (5 b e f o r e 7) = \frac{1}{9} + \frac{13}{18} . \frac{1}{9} + \frac{13}{18} . \frac{13}{18} . \frac{1}{9} + . . . . \infty = \frac{1}{9} [\frac{1}{1 - \frac{13}{18}}] = \frac{2}{5} .

Question 142. A box contains 24 identical balls of which 12 are white and 12 black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the

4^{t h}

time on the

7^{t h}

draw is

564
2732
532
12

Discuss Question

Answer: Option C. -> 532
:
C
In any trail, P(getting white ball) =

\frac{1}{2}

P(getting black ball) =

\frac{1}{2}

Now, required event will occur if in the first six trails 3 white balls are drawn in any one of the 3 trails from six. The remaining 3 trails must be kept reserved for black balls. This can happen
In

^{6} C_{3} \times^{3} C_{3} = 20

ways.

= 20 \times {(\frac{1}{2})}^{3} \times {(\frac{1}{2})}^{3} \times \frac{1}{2} = \frac{5}{32}

Question 143. The odds in favour of A solving a problem are 3 to 4 and the odds against B solving the same problem are 5 to 7. If they both try the problem, the probability that the problem is solved is:

4184
1621
521
14

Discuss Question

Answer: Option B. -> 1621
:
B

P (A) = \frac{3}{7}, P (B) = \frac{7}{12} P (¯ A) = \frac{4}{7}, P (¯ B) = \frac{5}{12} P (A \cup B) = 1 - P (¯ ¯¯¯¯¯¯¯¯¯¯¯¯ ¯ A \cup B) = 1 - P (¯ A \cap ¯ B) ∴ P (A \cup B) = 1 - \frac{4}{7} \times \frac{5}{12} = \frac{16}{21}

Question 144. Let `head` means 1 and `tail` means 2 and coefficients of the equation

a x^{2} + b x + c = 0

are chosen by tossing a fair coin. The probability that the roots of the equation are non-real, is equal to

Discuss Question

Answer: Option B. -> 78
:
B
a, b, c may be 1 or 2

a x^{2} + b x + c = 0

has non-real roots if

b^{2} - 4 a c < 0

\begin{matrix} b & 1 & 2 (a, c) & (1, 1), (1, 2), (2, 1), (2, 2) & (1, 2), (2, 1) (2, 2) \end{matrix} = 7

Hence required probabilidy

= \frac{7}{2^{3}} = \frac{7}{8} .

Question 145. The probability that a certain beginner at golf gets good shot if he uses correct club is

\frac{1}{3},

and the probability of a good shot with an incorrect club is

\frac{1}{4} .

In his bag there are 5 different clubs only one of which is correct for the good shot. If he chooses a club at random and take a stroke, the probability that he gets a good shot is

Discuss Question

Answer: Option C. -> 415
:
C
P( getting correct club )

= \frac{1}{5}

Therefore P(hitting good shot by correct club)

= \frac{1}{3} \times \frac{1}{5} = \frac{1}{15}

P(getting wrong club)

= \frac{4}{5}

P(hitting good shot with wrong club)

= \frac{4}{5} \times \frac{1}{4} = \frac{1}{5} .

Therefore P(hitting good shot)

= \frac{1}{15} + \frac{1}{5} = \frac{4}{15} .

Question 146. Sixteen players

P_{1} . P_{2} . \dots \dots \dots P_{16}

play in a tournament. They are divided into eight pairs at random.From each pair a winner is decided on the basis of a game played between the two players of the pair.Assuming that all the players are of equal strength, the probability that exactly one of the two players

P_{1}

and

P_{2}

is among the eight winners is

415
715
815
1730

Discuss Question

Answer: Option C. -> 815
:
C
Let

E_{1} (E_{2})

denote the event that

P_{1}

and

P_{2}

are paired (not paired) together and let A denote the event that one of two players

P_{1}

and

P_{2}

is among the winners.
Since, P1 can be paired with any of the remaining 15 players.
We have,

P (E_{1}) = \frac{1}{15}

and

P (E_{2}) = 1 - P (E_{1}) = 1 - \frac{1}{15} = \frac{14}{15}

In case

E_{1}

occurs, it is certain that one of

P_{1}

and

P_{2}

will be among the winners. In case

E_{2}

occurs, the probability that exactly of

P_{1}

and

P_{2}

is among the winners is

P {(P_{1} \cap_{2}) \cup (_{1} \cap P_{2})} = P (P_{1} \cap_{2}) + P (_{1} \cap P_{2}) = P (P_{1}) P (_{2}) + P (_{1}) P (P_{2}) = (\frac{1}{2}) (1 - \frac{1}{2}) + (1 - \frac{1}{2}) (\frac{1}{2}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}

ie,

P (\frac{A}{E_{1}}) = 1

and

P (\frac{A}{E_{2}}) = \frac{1}{2}

By the total probability Rule,

P (A) = P (E_{1}) . P (\frac{A}{E_{1}}) + P (E_{1}) . P (\frac{A}{E_{2}}) = \frac{1}{15} (1) + \frac{14}{15} (\frac{1}{2}) = \frac{8}{15}

Question 147. Two players A and B throw a die alternately for a prize of Rs 11, which is to be won by a player who first throws a six. If A starts the game, their respective expectations are

Rs 6; Rs 5
Rs 7; Rs 4
Rs 5.50; Rs 5.50
Rs 5.75; Rs 5.25

Discuss Question

Answer: Option A. -> Rs 6; Rs 5
:
A
Probability through a six =

\frac{1}{6}

P (A) = \frac{1}{6}, P (¯ A) = \frac{5}{6}, P (B) = \frac{1}{6}, P (¯ B) = \frac{5}{6}

Probability of A winning

= P (A) + P (¯ A) P (¯ B) P (A) + P (¯ A) P (¯ B) P (¯ A) P (¯ B) P (A) + . . .

= \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + . . .

= \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{6}{11}

Probability of B winning

= 1 - \frac{6}{11} = \frac{5}{11}

∴

Expectations of A and B are

\frac{6}{11} \times 11 =

Rs 6 and

\frac{5}{11} \times 11

=Rs 5

Question 148. A five digit number is formed with digits 0. 1. 2. 3. 4 without repetition. A number is selected at random, then the probability that it is divisible by 4 is

Discuss Question

Answer: Option B. -> 516
:
B
The number formed is divisible by 4 if the last two digits are 04,40,34,32,20,12.
Therefore total number of favorable ways = 3! + 3! +4 + 4+ 3! + 4 = 30 (This is the sum of number of ways in which the first two digits can be formed)
Total numbers that can be formed = 5! – 4! = 96 ( Number of ways of arranging 5 digits - Number of ways in which zero comes as the first digit)
Therefore required probability

= \frac{30}{96} = \frac{5}{16} .

Question 149. A coin whose faces marked 2 and 3 is thrown 5 times, then chance of obtaining a total of 12 is

Discuss Question

Answer: Option A. -> 516
:
A
It is given that the coin has faces 2 and 3, tossed five times.
To get sum 12 we need 2,2,2,3,3 in any combination.
So, in the 5 throws, there need to be 3 2's. Automatically other 2 will be 3's since there is no other possibility.
So, it boils down to choosing 3 from 5.
Therefore P(getting sum 12)

= 5 C_{3} {(\frac{1}{2})}^{5} = \frac{5}{16} .

Question 150. A bag contains 3 white, 3 black and 2 red balls. One by one, three balls are drawn without replacing them. Then the probability that the third ball is red , is given by

Discuss Question

Answer: Option C. -> 14
:
C
Number of white balls = 3, black balls = 3 red balls = 2
Since drawn balls are not replaced, for third ball to red, we have following patterns: