Question

A small deck of 5 cards consists of 3 red Jards and 2 green cards.
Draw 7 times with replacement.
Find the probability that a red card is drawn exactly 3 times.

Answer 1

Answer:

Approximately $0.19$.

Step-by-step explanation:

The cards were drawn with replacement. Therefore, the probability of drawing a red card will be $(3 / 5)$ at every one of the seven draws. The probability of not drawing a red card would be $(1 - (3/5)) = (2/5)$.

One way to get three red cards out of the seven draws would be $\verb!red!\; \verb!red!\; \verb!red!\; \verb!green!\; \verb!green!\; \verb!green!\; \verb!green!$. That corresponds to a probability of $(3/5)^3\, (2/5)^{7-3}$.

However, there are many other ways to get three red cards during the seven draws. For example:

• $\verb!red!\; \verb!red!\; \verb!green!\; \verb!red!\; \verb!green!\; \verb!green!\; \verb!green!$,
• $\verb!red!\; \verb!red!\; \verb!green!\; \verb!green!\; \verb!red!\; \verb!green!\; \verb!green!$.
• $\cdots$.
• $\verb!green!\; \verb!green!\; \verb!green!\; \verb!green!\; \verb!red!\; \verb!red!\; \verb!red!$.

There are $\displaystyle {\genfrac{(}{)}{0}{0}{7}{3}} = 35$ such ways in total to choose three draws out of seven draws.

These ways are mutually-exclusive. Each of these ways has a probability of $(3/5)^3\, (2/5)^{7-3}$. The probability for getting a red card on exactly three out of the seven draws would be:

$\displaystyle {\genfrac{(}{)}{0}{0}{7}{3}} \, (3/5)^3\, (1 - (3/5))^{7-3} \approx 0.19$.