JIPMAT 2023 (QA) - <p>The probabilities that A, B, and D can solve a problem independently are 1/3, 1/3, and 1/4 respectively. The probability that only two of them are able to solve the problem is:</p> | PYQs + Solutions | AfterBoards
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JIPMAT 2023 (QA) PYQs

JIPMAT 2023

Modern Math
>
Probability

Easy

The probabilities that A, B, and D can solve a problem independently are 1/3, 1/3, and 1/4 respectively. The probability that only two of them are able to solve the problem is:

Correct Option: 2
1) Let's first identify given probabilities:\newline - P(A)=13P(A) = \frac{1}{3}
- P(B)=13P(B) = \frac{1}{3}
- P(D)=14P(D) = \frac{1}{4}
2) For only two of them to solve:\newline We need to find:\newline - P(AB but not D)+P(AD but not B)+P(BD but not A)P(\text{AB but not D}) + P(\text{AD but not B}) + P(\text{BD but not A})
3) For AB but not D:\newline = P(A)×P(B)×(1P(D))P(A) × P(B) × (1-P(D))
= 13×13×34\frac{1}{3} × \frac{1}{3} × \frac{3}{4}
= 112\frac{1}{12}
4) For AD but not B:\newline = P(A)×P(D)×(1P(B))P(A) × P(D) × (1-P(B))
= 13×14×23\frac{1}{3} × \frac{1}{4} × \frac{2}{3}
= 118\frac{1}{18}
5) For BD but not A:\newline = P(B)×P(D)×(1P(A))P(B) × P(D) × (1-P(A))
= 13×14×23\frac{1}{3} × \frac{1}{4} × \frac{2}{3}
= 118\frac{1}{18}
6) Total probability:
= 112+118+118\frac{1}{12} + \frac{1}{18} + \frac{1}{18}
= 336+236+236\frac{3}{36} + \frac{2}{36} + \frac{2}{36}
= 736\frac{7}{36}
Therefore, the probability is 736\frac{7}{36}.

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