Question

A store has six different fitness magazines and three different news magazines. If a customer buys four magazines at random, what is the probability that the customer will buy three fitness magazines and one news mgazine?

• A. $\frac{2}{7}$
• B. $\frac{3}{7}$
• C. $\frac{4}{7}$
• D. $\frac{5}{7}$

Answer 1

The correct option is A $\frac{2}{7}$

Given: Total magazines $=9$
Fitness magazines $=6$
News magazines $=3$

Step 1: Number of ways to select three fitness magazines out of six is ${}^6{\text{C}}_3$.

$\Rightarrow {}^6{\text{C}}_3=\frac{6\times 5\times 4}{3\times 2}$

$=20$

Step 2: Number of ways to select one news magazine out of three is ${}^3{\text{C}}_1$.

$\Rightarrow {}^3{\text{C}}_1=3$

Step 3: Number of ways to select four (three fitness and one news) magazines out of ten is ${}^{10}{\text{C}}_4$.

$\Rightarrow {}^{10}{\text{C}}_4=\frac{10\times 9\times 8\times 7}{4\times 3\times 2}$

$=210$

Step 3: Probability to buy three fitness magazines and one news magazine

$=\frac{{}^6{\text{C}}_3\times {}^3{\text{C}}_1}{{}^{10}{\text{C}}_4}$

$=\frac{20\times 3}{210}$

$=\frac{2}{7}$