A bag contains 5 yellow and 4 green marbles. Three marbles are selected at random from the bag, without replacement.
(a) Show that the probability that exactly one of the marbles is yellow is \(\frac{5}{14}\).
The random variable \(X\) is the number of yellow marbles selected.
(b) Draw up the probability distribution table for \(X\).
(c) Find \(E(X)\).
Solution
(a) To find the probability that exactly one marble is yellow, consider the arrangements: YGG, GYG, GGY. The probability for each arrangement is:
\(\frac{5}{9} \times \frac{4}{8} \times \frac{3}{7}\)
Multiply by 3 for the three arrangements:
\(3 \times \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} = \frac{180}{504} = \frac{5}{14}\)
(b) The probability distribution table for \(X\) is:
| \(X\) | 0 | 1 | 2 | 3 |
|---|
| \(P(X)\) | \(\frac{1}{21}\) | \(\frac{5}{14}\) | \(\frac{10}{21}\) | \(\frac{5}{42}\) |
(c) The expected value \(E(X)\) is calculated as:
\(E(X) = 0 \times \frac{1}{21} + 1 \times \frac{5}{14} + 2 \times \frac{10}{21} + 3 \times \frac{5}{42} = 1.67\)
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