Question

A box consists of 5 yellow, 12 red and 8 blue balls. If 5 balls are drawn from this box one after the other without replacement, find the probability that the 5 balls are all yellow balls.

(a) \(\frac{5}{144}\)

(b) \(\frac{6}{321}\)

(c) \(\frac{4}{67}\)

(d) \(\frac{1}{231}\)

I had been asked this question in semester exam.

My query is from Multiplication Theorem on Probability topic in portion Discrete Probability of Discrete Mathematics

Answer 1

Correct option is (a) \(\frac{5}{144}\)

Easiest explanation: The total number of the balls in the box is 25. Let events Y: drawing black balls,

R: drawing red balls, B: drawing green balls. Now the balls are drawn without replacement. For the first draw, there are 25 balls to choose from, for the second draw it is 25 − 1 = 24 and 23 for the third draw. Then, the probability that the three balls are all yellow = P(Y1) P(Y2 | Y1) P(Y3 | Y1 ∩ Y2) = \(\frac{5}{24} * \frac{12}{24} * \frac{8}{24} = \frac{5}{144}\).