A charity organization's steering board consists of 6 donors and 4 volunteers. Two board m… — GMAT

A charity organization's steering board consists of $6$ donors and $4$ volunteers. Two board members are selected at random, one after another without replacement, to attend a national conference. What is the probability that the first member selected is a donor and the second member selected is a volunteer?

A
$\frac{6}{25}$
$\frac{4}{15}$ Answer
C
$\frac{1}{3}$
D
$\frac{8}{15}$
E
$\frac{12}{25}$

Answer

The probability that the first member selected is a donor and the second member selected is a volunteer is

\frac{4}{15}

The correct probability is calculated by multiplying the probability of the first event by the conditional probability of the second event given that the first event occurred. The probability of choosing a donor first is

\frac{6}{10}

. Since the selection is made without replacement, there are

9

total members remaining for the second draw,

4

of whom are volunteers. Therefore, the probability of selecting a volunteer second is

\frac{4}{9}

. The overall probability is

\frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15}

Step-by-Step Solution

Calculate the probability of selecting a donor on the first choice.

Since there are

6

donors out of

10

total board members,

P(\text{1st Donor}) = \frac{6}{10} = \frac{3}{5}

The sample space initially contains

10

members,

6

of whom are donors.

Calculate the conditional probability of selecting a volunteer on the second choice given that a donor was chosen first.

After one donor is selected,

9

members remain,

4

of whom are volunteers. Thus,

P(\text{2nd Volunteer} \mid \text{1st Donor}) = \frac{4}{9}

The selection is made without replacement, reducing both the total number of members in the pool and the sample space size.

Multiply the sequential probabilities for dependent events.

P(\text{1st Donor and 2nd Volunteer}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15}

For dependent events

A

and

B

, the joint probability is

P(A \cap B) = P(A) \times P(B \mid A)

Key Concept

Probability of Dependent Events (Sampling Without Replacement)

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