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 box, what is the probability that it is a silver coin.

• A. $0.3$
• B. $0.4$
  
• C. $0.5$
• D. $0.6$

Answer 1

The correct option is A $0.3$

Box $I:2$ Silver coins $+4$ gold coins

Box $II:4$ Silver coins $+3$ gold coins

Let $E_1$ be the event of selecting a gold coin from box $I$

and $E_2$ be the event of selecting a gold coin from box $II.$

$A:$ Probability of gold coin

$\Rightarrow P\left(E_1\right)={}^6{C}_2$ $P\left(E_2\right)={}^7{C}_4$

$\Rightarrow P\left(\frac{A}{E_1}\right)=P\left(\frac{A}{E_2}\right)=\frac{1}{2}$

$\Rightarrow P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1\right)P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)P\left(E_2\right)P\left(\frac{A}{E_2}\right)}$

$=\frac{{}^6{C}_2\times \frac{1}{2}}{{}^6{C}_2\times \frac{1}{2}+{}^7{C}_4\times \frac{1}{2}}$

$=\frac{7.5}{7.5+17.5}$

$=0.3$

Hence, the answer is $0.3$