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Chapter 9: Problem 33

Find the probability for the experiment of drawing two marbles at random (without replacement) from a bag containing one green, two yellow, and three red marbles. Neither marble is yellow.

Short Answer

Expert verified
The probability of drawing two marbles at random (without replacement) from a bag containing one green, two yellow, and three red marbles and neither marble being yellow is 0.4

Step by step solution

01

Total number of marbles

The total number of marbles in the bag is \(1 (\text{green}) + 2 (\text{yellow}) + 3 (\text{red}) = 6\).
02

Total ways of drawing 2 marbles without replacement

The total number of ways of drawing 2 marbles without replacement from 6 is represented by the combination \(C(6, 2)\). In general, \(C(n, k)\) is given by \(\frac{n!}{k!(n-k)!}\), where \(n!\) is the factorial of \(n\), so \(C(6, 2) = \frac{6!}{2!(6-2)!} = 15\).
03

Number of favorable outcomes

The favorable outcome is that neither marble is yellow. Hence, the only options available are green and red marbles. The number of green and red marbles is \(1 (\text{green}) + 3 (\text{red}) = 4\). The number of ways of drawing 2 marbles from these 4 is represented by the combination \(C(4, 2)\), which evaluates to \(\frac{4!}{2!(4-2)!} = 6\).
04

Calculation of Probability

The probability of an event is given by the ratio of the number of favorable outcomes to the total number of outcomes. Hence, in this case, the probability, \(P\), is \(P = \frac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}} = \frac{C(4, 2)}{C(6, 2)} = \frac{6}{15} = 0.4\)

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