Question

Find the probability of getting 4 heads in 6 tosses of fair coin

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

ASSUMPTIONS

• The coin is completely fair

• The coin lands only on one flat face or the other side, a Head or a Tail.

• The coin is tossed fairly.

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{a \: coin \: is \: tossed \: 6 \: times} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{probability \: of \: getting \: 4 \: heads}  \end{cases}\end{gathered}\end{gathered}$

Now,

Let p represent the probability of getting head when a coin is tossed and q represents the probability of getting tail when a coin is tossed.

$\rm: \implies \: p \: = \: \dfrac{1}{2} \: \: and \: \: q \: = \: \dfrac{1}{2}$

Here, coin is tossed 6 times

$\rm: \implies \: n \: = \: 6$

As we know,

By Binomial distribution,

The probability of exactly k successes in n trials with probability p of success and q of failure in any trial, is:

$\rm: \implies \: P(k) \: = \: \:^n C_k \: {p}^{k} \: {q}^{n - k}$

where,

$\boxed{\sf \:^{n}C_k=\dfrac{n!}{(n-k)!\times k!}}$

So, required probability of getting 4 heads when coin is tossed 6 times is given by

$\rm: \implies \: P(4) \: = \: \:^6 C_4 \times {(\dfrac{1}{2}) }^{4} \times {(\dfrac{1}{2} )}^{2}$

$\rm: \implies \: P(4) \: = \dfrac{6!}{4! \times 2!} \times \dfrac{1}{16} \times \dfrac{1}{4}$

$\rm: \implies \: P(4) \: = \: \dfrac{6 \times 5 \times 4!}{4! \times 2 \times 1} \times \dfrac{1}{64}$

$\rm: \implies \: P(4) \: = \: \dfrac{15}{64}$