Red vs Black

Probability Level 5

A bag contains 3 red and 4 black balls. Abhineet, randomly decides to toss a coin or a die with probabilities 1 3 \frac{1}{3} and 2 3 \frac{2}{3} respectively.

If a die is tossed, and the number obtained is a multiple of 3, he does not remove nor add extra balls into the bag. If the number is not a multiple of 3, he picks a ball randomly out of the bag, and replaces the ball along with an additional ball of the same colour.

If a coin is tossed, and a head is obtained he picks a ball randomly of the bag and discards it. If a tail is obtained, he discards one of the black balls.

He does this process once.

After the process is done, he picks a ball out of the bag randomly.

Given that the probability of picking a red ball is A B \dfrac{A}{B} , where A A and B B are coprime positive integers.

Find B − A B - A .


The answer is 47.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

The probability of picking a red ball in the first branch ( P 1 ) \left( P_{1} \right) = 1 3 × 1 2 × ( 2 6 × 3 7 + 3 6 × 4 7 + 0 × 2 6 + 1 × 3 6 ) \dfrac{1}{3} \times \dfrac{1}{2} \times \left( \dfrac{2}{6} \times \dfrac{3}{7} + \dfrac{3}{6} \times \dfrac{4}{7} + 0 \times \dfrac{2}{6} + 1 \times \dfrac{3}{6} \right) = 1 6 × ( 6 42 + 12 42 + 21 42 ) \dfrac{1}{6} \times \left( \dfrac{6}{42} + \dfrac{12}{42} + \dfrac{21}{42} \right) = 1 6 × ( 39 42 ) \dfrac{1}{6} \times \left( \dfrac{39}{42} \right) = 13 84 \dfrac{13}{84}

The probability of picking a red ball in the second branch ( P 2 ) = 2 3 × ( 1 3 × 3 7 + 2 3 × 3 7 × 4 8 + 2 3 × 4 7 × 3 8 ) \left( P_{2} \right) = \dfrac{2}{3} \times \left( \dfrac{1}{3} \times \dfrac{3}{7} + \dfrac{2}{3} \times \dfrac{3}{7} \times \dfrac{4}{8} + \dfrac{2}{3} \times \dfrac{4}{7} \times \dfrac{3}{8} \right) = 2 3 × ( 1 7 + 2 3 × 24 56 ) \dfrac{2}{3} \times \left( \dfrac{1}{7} + \dfrac{2}{3} \times \dfrac{24}{56} \right) = 2 3 × ( 1 7 + 2 7 ) \dfrac{2}{3} \times \left( \dfrac{1}{7} + \dfrac{2}{7} \right) = 2 3 × 3 7 \dfrac{2}{3} \times \dfrac{3}{7} = 2 7 \dfrac{2}{7}

Total probability P = P 1 + P 2 P = P_{1} + P_{2}
∴ P = 13 84 + 2 7 = 13 84 + 24 84 = 37 84 \therefore P = \dfrac{13}{84} + \dfrac{2}{7} = \dfrac{13}{84} + \dfrac{24}{84} = \dfrac{37}{84}

Comparing, we get
A = 37 A = 37
B = 84 B = 84

B − A = 84 − 37 = 47 B - A = 84 - 37 = 47

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution!

A Former Brilliant Member - 5 years, 5 months ago

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Thanks. Using excel seemed the way to go with the branching.

A Former Brilliant Member - 5 years, 5 months ago

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Yep, and it does make it easy for the reader as well!

A Former Brilliant Member - 5 years, 5 months ago

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