A problem is given to 3 students A, B and C w | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$,and $P(C) = \frac{1}{4}$ be the probabilities of students $A, B$,and $C$ solving the problem respect

Vedclass · Accepted Answer

$\frac{11}{24}$

Vedclass · Answer

(B) Let $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$,and $P(C) = \frac{1}{4}$ be the probabilities of students $A, B$,and $C$ solving the problem respectively.Then,the probabilities of them not solving the problem are $P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}$,$P(\bar{B}) = 1 - \frac{1}{3} = \frac{2}{3}$,and $P(\bar{C}) = 1 - \frac{1}{4} = \frac{3}{4}$.The probability that exactly one of them solves the problem is given by:$P(	ext{Exactly one}) = P(A)P(\bar{B})P(\bar{C}) + P(\bar{A})P(B)P(\bar{C}) + P(\bar{A})P(\bar{B})P(C)$$= (\frac{1}{2} 	imes \frac{2}{3} 	imes \frac{3}{4}) + (\frac{1}{2} 	imes \frac{1}{3} 	imes \frac{3}{4}) + (\frac{1}{2} 	imes \frac{2}{3} 	imes \frac{1}{4})$$= \frac{6}{24} + \frac{3}{24} + \frac{2}{24} = \frac{11}{24}$.

$A$ problem is given to $3$ students $A, B$ and $C$ whose chances of solving it are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively. Then,the probability of the problem being solved by exactly one of them,if all the three try independently,is

Similar Questions

Six faces of an unbiased die are numbered with $2, 3, 5, 7, 11$ and $13$. If two such dice are thrown, then the probability that the sum on the uppermost faces of the dice is an odd number is

Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined: $P(A) = 0.5$,$P(B) = 0.7$,$P(A \cap B) = 0.6$.

In a single throw of two dice,what is the probability of obtaining an odd number as the sum of the numbers on the two dice?

$A$ die is tossed thrice. Find the probability of getting an odd number at least once.

When two dice are rolled,what is the probability that at least one face shows a six?

Vedclass Test Series

Exam Paper Generator

Online Exam Module