Question ID: #829
Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A. Then a ball is randomly drawn from the bag A. If the probability, that the ball drawn is white, is $\frac{p}{q}$, $gcd(p,q)=1$, then $p+q$ is equal to:
- (1) 22
- (2) 23
- (3) 24
- (4) 21
Solution:
Let $E_1$ be the event that a White ball is transferred from Bag B to Bag A.
Let $E_2$ be the event that a Black ball is transferred from Bag B to Bag A.
Let $W$ be the event that a White ball is drawn from Bag A.
Probability of transferring White from B ($P(E_1)$):
$$ P(E_1) = \frac{6}{10} = \frac{3}{5} $$
Probability of transferring Black from B ($P(E_2)$):
$$ P(E_2) = \frac{4}{10} = \frac{2}{5} $$
If $E_1$ occurs, Bag A has $9+1=10$ White and 8 Black balls (Total 18).
$$ P(W|E_1) = \frac{10}{18} $$
If $E_2$ occurs, Bag A has 9 White and $8+1=9$ Black balls (Total 18).
$$ P(W|E_2) = \frac{9}{18} $$
Using the Law of Total Probability:
$$ P(W) = P(E_1)P(W|E_1) + P(E_2)P(W|E_2) $$
$$ P(W) = \left( \frac{3}{5} \times \frac{10}{18} \right) + \left( \frac{2}{5} \times \frac{9}{18} \right) $$
$$ P(W) = \frac{30}{90} + \frac{18}{90} = \frac{48}{90} $$
Simplifying the fraction:
$$ \frac{48}{90} = \frac{8}{15} $$
Here, $p = 8$ and $q = 15$. $\text{gcd}(8, 15) = 1$.
The value of $p + q$:
$$ p + q = 8 + 15 = 23 $$
Ans. (2)
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