A bag contains 4 tickets numbered 1, 2, 3, 4 and another bag contains 6 tickets numbered 2, 4, 6, 7, 8, 9. One bag is chosen and a ticket is drawn. The probability that the ticket bears the number 4, is equal to
Answer & explanation
Correct answer: option 2
$E_1$ : First bag is chosen, $P\left(E_1\right)=\frac{1}{2}$
$E_2$ : Second bag is chosen, $P\left(E_2\right)=\frac{1}{2}$
A : Drawn number is 4
Now, $P(A)=P\left(E_1\right) . P\left(A / E_1\right)+P\left(E_2\right) . P\left(A / E_2\right)$
$=\frac{1}{2} . \frac{1}{4}+\frac{1}{2} . \frac{1}{6}=\frac{5}{24}$