A sports store owner conducts a game 'weekend-surprise' every Friday for his customers. He fills…
A sports store owner conducts a game 'weekend-surprise' every Friday for his customers.
He fills two bags with cricket balls of red and white colours. The first bag has 4 white and 4 red balls while the second bag contains 3 white and 5 red balls.
The rules of the game are:
- The customer will be blind folded.
- Two balls have to be transferred from the first bag to the second bag one after another without replacement, and then one ball has to be drawn out from the second bag.
- The colours of the three balls (two balls transferred from the first bag and one ball drawn from the second bag) are considered.
- If all the three balls are of the same colour, the customer wins a surprise gift.
What is the probability that a customer can win the surprise gift?
Answer
Answer
AIWritten by AI (gemini-2.5-pro) - it can contain mistakes.
Bag 1 has 4 white and 4 red balls (8 total); Bag 2 has 3 white and 5 red balls (8 total).
A customer wins if all 3 balls are of the same colour.
Case 1 (All 3 White):
$P(W_1 W_2) = \frac{4}{8} \times \frac{3}{7} = \frac{3}{14}$.
After transfer, Bag 2 has 5 white and 5 red balls (10 total).
$P(W_3 \mid W_1 W_2) = \frac{5}{10} = \frac{1}{2}$.
$P(\text{all White}) = \frac{3}{14} \times \frac{1}{2} = \frac{3}{28}$.
Case 2 (All 3 Red):
$P(R_1 R_2) = \frac{4}{8} \times \frac{3}{7} = \frac{3}{14}$.
After transfer, Bag 2 has 3 white and 7 red balls (10 total).
$P(R_3 \mid R_1 R_2) = \frac{7}{10}$.
$P(\text{all Red}) = \frac{3}{14} \times \frac{7}{10} = \frac{3}{20}$.
Total probability:
$P(\text{win}) = \frac{3}{28} + \frac{3}{20} = \frac{15 + 21}{140} = \frac{36}{140} = \frac{9}{35}$.
From ISC 2026 Mathematics Paper 1, question 10(i).