Question

A coin is so weighted such that the probability of it showing H(head) is $\frac{2}{3}$and that of T(Tail)is $\frac{1}{3}.$ When it is tossed. If head appears, then a number from the first 9 naturals is selected at random, otherwise a number from 1, 2, 3, 4, 5 will be selected. Let E be the event of getting an even number, then

• A. $P\left(\frac{E}{H}\right)=\frac{4}{9}$
  
• B. $P\left(\frac{E}{T}\right)=\frac{2}{5}$
  
• C. $P\left(E\right)=\frac{58}{135}$
  
• D. $P\left(\frac{H}{E}\right)=\frac{20}{29}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A

$P\left(\frac{E}{H}\right)=\frac{4}{9}$

B

$P\left(\frac{E}{T}\right)=\frac{2}{5}$

C

$P\left(E\right)=\frac{58}{135}$

D

$P\left(\frac{H}{E}\right)=\frac{20}{29}$

Since there are 4 even number among 1 to 9, it follow that
$P\left(\frac{E}{H}\right)=\frac{4}{9}$
Again, since there are 2 even number among 1 to 5, it follow that
$P\left(\frac{E}{T}\right)=\frac{2}{5}$
Therefore (A) and (B) are correct, now
$E=(H\cup T)\cap E=(H\cap E)\cup (T\cap E)$
This implies
$P\left(E\right)=P(H\cap E)+P(T\cap E)=P\left(H\right)P\left(\frac{E}{H}\right)+P\left(T\right)P\left(\frac{E}{T}\right)=\frac{2}{3}\times \frac{4}{9}+\frac{1}{3}\times \frac{2}{5}=\frac{58}{135}$
By Baye’s theorem
$P\left(\frac{H}{E}\right)=\frac{P\left(H\right)P\left(\frac{E}{H}\right)}{P\left(H\right)P\left(\frac{E}{H}\right)+P\left(T\right)P\left(\frac{E}{T}\right)}=\frac{\frac{2}{3}\times \frac{4}{9}}{\frac{2}{3}\times \frac{4}{9}+\frac{1}{3}\times \frac{2}{5}}=\frac{20}{29}$