A box is selected: box $A$ with probability $p$, and box $B$ with probability $1-p$. Box \(A

Question

A box is selected: box $A$ with probability $p$, and box $B$ with probability $1-p$. Box $A$ contains 3 red and 1 blue counter. Box $B$ contains 1 red and 3 blue counters. One counter is drawn from the selected box. Show that $P(R)=\frac{1+2p}{4}$. Find $P(A\mid R)$ in terms of $p$. Find the values of $p$ for which $P(A\mid R)>\frac12$.

Accepted Answer

(a) $P(R)=\frac{1+2p}{4}$. (b) $P(A\mid R)=\frac{3p}{1+2p}$. (c) $\frac14<p\le1$.

Probability

Probability

Solution:

Solution locked