Multiple choice

Bag $A$ contains three red and four white Balls; bag $B$ contains two red and three white balls. If one ball is drawn from bag $A$ and two balls from bag $B$, find the probability that: (i) One ball is red and two balls are white; (ii) All the three balls are of the same colour.

  1. i) $\cfrac {33}{70}$ and ii) $\cfrac {3}{14}$
  2. i) $\cfrac {22}{31}$ and ii) $\cfrac {1}{3}$
  3. i) $\cfrac 12$ and ii) $\cfrac 12$
  4. i) $\cfrac {8}{25}$ and ii) $\cfrac {2}{7}$
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A Correct answer
Explanation

Bag A: 3R, 4W (Total 7). Bag B: 2R, 3W (Total 5). Drawing 1 from A, 2 from B. (i) 1R from A, 2W from B OR 1W from A, 1R+1W from B. P(i) = (3/7 * 3/5 * 2/4) + (4/7 * (2/5 * 3/4 + 3/5 * 2/4)) = 18/140 + 4/7 * 12/20 = 18/140 + 48/140 = 66/140 = 33/70. (ii) All red: 3/7 * 2/5 * 1/4 = 6/140. All white: 4/7 * 3/5 * 2/4 = 24/140. Total = 30/140 = 3/14.

AI explanation

For the first event, the probability is the sum of drawing one red from A and two white from B, and one white from A and two red from B; this is (3/7 multiplied by 3C2/5C2) plus (4/7 multiplied by 2C2/5C2), equaling 33/70. For the second event, all three balls can only be the same color if they are all white, making the probability 4/7 multiplied by 3C2/5C2, which equals 3/14.

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