Question

Box A contains $3$ red and $2$ blue marbles while box $B$ contains $2$ red and $8$ blue marbles. A fair coin is tossed. If the coin turns up heads, a marbles is drawn from $A$, if it turns up tails, a marble is drawn from bag $B$. the probability that a red marble is chosen,is

• A. $\frac{1}{5}$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{2}{5}$

A] Event A = { o red marbles is chosen}

$B_1=$ {Box 1 is chosen}

$B_2=$ {Box 2 is chosen}

By formula for total probability

$P\left(A\right)=P\left(A/B_1\right)P\left(B_1\right)+P\left(A_1{B}_1\right)P\left(B_2\right)$

$P\left(B_1\right)=P\left(B_2\right)=\frac{1}{2},P\left(A_1{B}_1\right)=\frac{3}{5}$

$P\left(A_1{B}_2\right)=\frac{2}{10}=\frac{1}{5}$

So, $P\left(A\right)=\frac{1}{2}\times \frac{3}{5}+\frac{1}{2}\times \frac{1}{5}=\frac{1}{10}=\frac{2}{5}$