Three students A, B and C are running a race. A and B have the same probability of winning and each is twice…

Three students A, B and C are running a race. A and B have the same probability of winning and each is twice likely to win as C. Then, the probability that B or C wins is equal to (assuming there are no ties)
  1. $\frac{2}{5}$
  2. $\frac{3}{5}$
  3. $\frac{3}{7}$
  4. $\frac{2}{7}$

Solution

Let winning probability of C be P(C) = p $\begin{aligned} & \therefore \quad P(A)=P(B)=2 p \\ & \because \quad P(A)+P(B)+P(C)=1 \\ & 2 p+2 p+p=1 \\ & \Rightarrow \quad 5 p=1 \\ & p=1 / 5 \\ & \end{aligned}$ $\therefore$ Probability that $B$ or $C$ win $=P(B)+P(C)$ $=2 p+p[\because B$ and $C$ are mutually exclusive events $]$ $ =3 p=\frac{3}{5} $

Asked in: AP EAMCET 2021 (25 Aug Shift 2)

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