Skip to main contentSkip to solution

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:

Solution

Correct Option: 2
  1. Let's first identify given probabilities:
  • P(A)=13P(A) = \frac{1}{3}
    • P(B)=13P(B) = \frac{1}{3}
    • P(D)=14P(D) = \frac{1}{4}
  1. For only two of them to solve:

We need to find:

  • 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})
  1. For AB but not D:

= 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}

  1. For AD but not B:

= 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}

  1. For BD but not A:

= 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}

  1. 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}.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question