A box has 4 green, 8 blue, and 3 red pens. A student picks up a pen at random, checks its color and replaces it in the box. He repeats this process 3 times. The probability that at least one pen picked was red is:

Question

The distribution function \( F(X) \) of a discrete random variable \( X \) is given. Then \( P[X = 4] + P[X = 5] \):
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline F(X = x) & 0.2 & 0.37 & 0.48 & 0.62 & 0.85 & 1 \\ \hline \end{array} \]

Answer 1

A box contains 4 green, 8 blue, and 3 red pens, for a total of \( 4 + 8 + 3 = 15 \) pens. The probability of selecting a red pen in a single trial is \( P(\text{red}) = \frac{3}{15} = \frac{1}{5} \). Consequently, the probability of not selecting a red pen is \( P(\text{not red}) = 1 - P(\text{red}) = 1 - \frac{1}{5} = \frac{4}{5} \). The probability of not selecting a red pen in 3 independent trials is \( \left(\frac{4}{5}\right)^3 = \frac{64}{125} \). Therefore, the probability of selecting at least one red pen in 3 trials is \( 1 - \frac{64}{125} = \frac{61}{125} \). The correct answer is (3).

A box has 4 green, 8 blue, and 3 red pens. A student picks up a pen at random, checks its color and replaces it in the box. He repeats this process 3 times. The probability that at least one pen picked was red is:

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The Correct Option is C

Solution and Explanation

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