Question

A box contains $2$ silver coins and $4$ gold coins and the second box contains $4$ silver coins and $3$ gold coins . If a coin is selected from one of the boxes , what is the probability that it is a silver coin?
A $0.3$
B $0.4$
C $0.5$
D $0.6$

Answer 1

In this question the total probability theorem is used as the probability of selecting one of the box is $\dfrac{1}{2}$ and the probability of silver coin from box $1$ and $2$ is $\dfrac{1}{3}$ and $\dfrac{4}{7}$ hence probability of silver coin $P(S) = P(1)P\left( {\dfrac{S}{1}} \right) + P(2)P\left( {\dfrac{S}{2}} \right)$

Complete step-by-step answer:
It is given in the question that
Box $1$ contains $2$ silver coins $+$ $4$ gold coins
Box $2$ contains $4$ silver coins $+$ $3$ gold coins
Hence selection of the two boxes is independent to each other ,
Therefore
Probability of selecting the Box $1$ is equal to $P(1) = \dfrac{1}{2}$
Probability of selecting the Box $2$ is equal to $P(2) = \dfrac{1}{2}$
In the Box $1$ there is $2$ silver coins $+$ $4$ gold coins hence the probability of silver coin from Box $1$ is
= $P\left( {\dfrac{S}{1}} \right) = \dfrac{2}{6} = \dfrac{1}{3}$
In the Box $2$ there is $4$ silver coins $+$ $3$ gold coins hence the probability of silver coin from Box $2$ is
= $P\left( {\dfrac{S}{2}} \right) = \dfrac{4}{7}$
Hence from the Total Probability theorem ,
For the probability coin is selected from one of the box and it is silver coin
$P(S) = P(1)P\left( {\dfrac{S}{1}} \right) + P(2)P\left( {\dfrac{S}{2}} \right)$
$P(S) = \dfrac{1}{2} \times \dfrac{1}{3} + \dfrac{1}{2} \times \dfrac{4}{7}$

$P(S) = \dfrac{1}{6} + \dfrac{4}{{14}}$
$P(S) = 0.16 + 0.28$
$P(S) = 0.45$
Probability of coin that is selected from one of the box and it is silver coin is $P(S) = 0.45$

So, the correct answer is “Option B”.

Note: As in this question the probability of coin is asked as it would be drawn from any of the boxes if it specifies the probability of coin drawn from Box $1$ then we have to use Bayes Theorem for calculating the probability .
Probability of any event always lies between $0$ to $1$ . If your answer comes apart from this then cross check it