Question

A bag contains five red marbles, two orange marbles, one yellow marble, and two green marbles. Two marbles are drawn from the bag. What is the approximate probability of choosing an orange marble and a green marble? 0. 02222 0. 04444 0. 08889 0. 13333.

Answer 1

Option C: 0.8889 is the probability of choosing an orange marble and a green marble.

Given that:

• Number of red marbles is 5

• Number of orange marbles is 2

• Number of yellow marbles is 1.

• Number of green marbles is 2.

Explanation and Calculations:

So there are in total 10 marbles.

Two marbles can be chosen in $^{10}C_2$ ways = 90/8

Now the favorable event is when one green marble and one orange marble  is chosen. The number of ways this can be done is $^{2}C_1 \times ^{2}C_1 = 4$

The resultant probability is calculated by:

$P(E) = \dfrac{Number \:of\: Favourable \:cases}{ Number\: of\: total \:cases}\\= \dfrac{4}{\frac{90}{2}}\\\\= \dfrac{8}{90}\\\\\approx 0.08889$

Thus the needed probability is 0.08889 given by option C.

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