# Problem of the Day

An unfair coin comes up heads 60% of the time and tails 40% of the time when it is tossed. What is the probability of getting exactly three heads on five tosses of this coin?

A.

B.

C.

D.

E.

[latexpage]

E. $\displaystyle \frac{6^3}{5^4}$

See the Solution

### Solution

[latexpage]

Consider one possible successful sequence: HHHTT. Each toss has a $\frac{3}{5}$ probability of being heads and a $\frac{2}{5}$ probability of being tails. Thus, this sequence will occur with probability $(\frac{3}{5})^3\cdot(\frac{2}{5})^2$. However, this is only one of many possible successful sequences. To determine the total number of sequences, note that there are 5 tosses and the two tails can be assigned to any of those five tosses in $_5C_2=10$ ways. Thus, the overall probability is $10\cdot(\frac{3}{5})^3\cdot(\frac{2}{5})^2$.

$10\cdot(\frac{3}{5})^3\cdot(\frac{2}{5})^2$

$=2\cdot5\cdot\frac{3^3}{5^3}\cdot\frac{2^2}{5^2}$

$=\frac{3^3}{5^2}\cdot\frac{2^3}{5^2}$

$=\frac{6^3}{5^4}$