Question

A coin is so biased that the probability of falling head when tossed is $\frac{1}{4}$ . If the coin is tossed $5$ times the probability of obtaining $2$ heads and $3$ tails, with heads occurring in succession is

• A. $\frac{5\times 3^3}{4^5}$
• B. $\frac{3^3}{5^4}$
• C. $\frac{3^3}{4^4}$
• D. $\frac{3^3}{4^5}$

Answer 1

The correct option is C $\frac{3^3}{4^4}$

$P\left(H\right)=\frac{1}{4}$
And
$P\left(T\right)=\frac{3}{4}$
Hence the required probability will be
$HHTTT$
$=\frac{1}{4}.\frac{1}{4}(\frac{3}{4}{)}^3$
$=\frac{3^3}{4^5}$.
Now consider $HHTTT$.
Since $HH$ has to be together, we consider $HH$ as one item.
Hence total number of arrangements of $4$ items in which $3$ are repeated are
$\frac{4!}{3!}$
$=4$.
Hence $HHTTT$ can be internally permuted in $4$ ways considering that the heads occur in succession.
Therefore the required probability will be
$=\frac{3^3}{4^5}.4$
$=\frac{3^3}{4^4}$.