Question-68

A box contains 4 red balls and 6 blue balls. Two balls are drawn one by one, without replacement. Let \(X\) be the number of red balls drawn. What is the probability that exactly 1 red ball is drawn, given that at least one red ball is drawn\(?\)

Choose the correct options.

• \(\dfrac{8}{15}\)

• \(\dfrac{1}{3}\)

• \(\dfrac{2}{3}\)

• \(\dfrac{4}{5}\)

Hint

We have to calculate \(P(X = 1 ~ | ~ X \geq 1)\), which is the conditional probability.

So, \(P(X = 1 ~ | ~ X \geq 1) = \dfrac{P(X = 1 \cap X \geq 1)}{P(X \geq 1)}\)

Answer

• \(\dfrac{8}{15}\)

• \(\dfrac{1}{3}\)

• \(\dfrac{2}{3}\)

• \(\dfrac{4}{5}\)

Solution

\(P(X = 1 ~ | ~ X \geq 1) = \dfrac{P(X = 1 \cap X \geq 1)}{P(X \geq 1)}\)

Now, \(P(X = 1) = \dfrac{^4C_1 \times ^6C_1}{^{10}C_2} = \dfrac{24}{45} = \dfrac{8}{15}\)

Next, calculate \(P(X \geq 1) = 1 - P(X < 1) = 1 - P(X = 0)\)

\(P(X = 0 ) = \dfrac{^4C_0 \times ^6C_2}{^{10}C_2} = \dfrac{15}{45} = \dfrac{1}{3}\)

So, \(P(X \geq 1 ) = 1 - P(X =0) = 1- \dfrac{1}{3} = \dfrac{2}{3}\)

Thus, the required probability:

\(P(X = 1 ~ | ~ X \geq 1) = \dfrac{P(X = 1 \cap X \geq 1)}{P(X \geq 1)} = \dfrac{8}{15} \times \dfrac{3}{2} = \dfrac{4}{5}\)