Question
Six coins are tossed simultaneously. Find the probability of getting.
3 heads.

Answer

Let p represents the probability of getting head in a toss of fair coin, so
$\text{p}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}$ [Since p + q = 1]
$\text{q}=\frac{1}{2}$
Let X denote the random variable representing the number heads in 6 tosses of coin. probability of getting r sixes in n tosses of a fair coin is given by,
$\text{P(X = r})=\text{ }^\text{n}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$3
$=\text{ }^6\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{6-\text{r}}\dots(1)$
Probability of getting 3 heads
$=\text{P(X}=3)$
$=\text{ }^6\text{c}_3\big(\frac{1}{2}\big)^3\big(\frac{1}{2}\big)^{6-3}$
$=\frac{6\times5\times4}{3\times2}\big(\frac{1}{2}\big)^3\big(\frac{1}{2}\big)^3$
$=\frac{20}{64}$
Probability of getting 3 heads $=\frac{20}{64}=\frac{5}{16}$

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