Multiple choice

Bag $A$ contains $6$ red and $4$ black balls and bag $B$ contains $4$ red and $6$ black balls. One ball is drawn at random from bag $B$ and placed in bag $A$. If one ball is now drawn from bag $A$. The probability that it is found to be red is

  1. $\dfrac {32}{55}$
  2. $\dfrac {33}{55}$
  3. $\dfrac {32}{63}$
  4. $\dfrac {25}{66}$
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A Correct answer
Explanation

There are two cases: drawing a red ball from B (prob 4/10) or a black ball from B (prob 6/10). If red is moved, Bag A becomes 7 red, 4 black (prob 7/11). If black is moved, Bag A becomes 6 red, 5 black (prob 6/11). Total probability = (4/10 * 7/11) + (6/10 * 6/11) = 28/110 + 36/110 = 64/110 = 32/55.

AI explanation

Using the law of total probability, we calculate the chance of drawing a red ball from bag A by considering the two scenarios for the transferred ball. The probability of transferring a red ball is 4/10, which leaves 7 red among 11 balls in bag A for a resulting probability of (4/10) times (7/11). The probability of transferring a black ball is 6/10, which leaves 6 red among 11 balls in bag A for a resulting probability of (6/10) times (6/11). Adding these gives (28/110) plus (36/110), which equals 64/110. The result is 32/55.

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