Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables X and Y respectively denote the number of blue and yellow balls. If X̄ and Ȳ are the means of X and Y respectively, then 7 X̄+4 Ȳ is equal to ___________.

Solution

Maths sir · 4 steps

Find the mean of X, and Y comes free, because every draw has exactly 3 balls, so Y = 3 - X.

1. Set up. There are C(9, 3) = 84 equally likely ways to draw 3 balls. If X = k blue balls are drawn, the other 3 - k are yellow, so Y = 3 - X every time.

   X (blue)	Y (yellow)	Ways	P(X)
   0	3	C(5, 0)C(4, 3) = 4	4/84
   1	2	C(5, 1)C(4, 2) = 30	30/84
   2	1	C(5, 2)C(4, 1) = 40	40/84
   3	0	C(5, 3)C(4, 0) = 10	10/84

2. Mean of X. X̄ = (0 × 4 + 1 × 30 + 2 × 40 + 3 × 10)/84 = 140/84 = 5/3. This matches the quick rule 3 × 5/9: each of the 3 draws is blue with probability 5/9.
3. Mean of Y. Since Y = 3 - X, the mean is Ȳ = 3 - 5/3 = 4/3, which is the same as 3 × 4/9.

   7X̄ + 4Ȳ = 7 × 5/3 + 4 × 4/3 = (35 + 16)/3 = 51/3

4. So. 7X̄ + 4Ȳ = 17.

The common slip: 16. Swapping the means, taking X̄ = 4/3 and Ȳ = 5/3, gives (28 + 20)/3 = 16, not 17. X counts blue balls, and there are 5 blue.

The idea to keep. When you draw n items from a group, the expected number of a type is n × (fraction of that type), with or without replacement. Check it against the full distribution once to trust it.

Answer from the NTA final answer key. Solution by PrepWiser.