Question

Assertion :If 10 coins are thrown simultaneously, then the probability of exactly 4 heads appearing is equal to the probability of exactly 6 heads appearing. Reason: ${}^n{\text{C}}_x{=}^n{\text{C}}_y.\Rightarrow$ either $x=y$ or $n=x+y$ and $P\left(H\right)=P\left(T\right)$ in a single toss

• A. Both Assertion & Reason are individually true & Reason is correct explanation of Assertion
• B. Both Assertion & Reason are individually true but Reason is not the ,correct (proper) explanation of Assertion
• C. Assertion is true but Reason is false
• D. Assertion is false but Reason is true

Answer 1

The correct option is A Both Assertion & Reason are individually true & Reason is correct explanation of Assertion

Let P(getting a head) = $p=\frac{1}{2}$

and let P(getting a tail)= $q=1-p=1-\frac{1}{2}=\frac{1}{2}$

As n = 10

The general form of binomial distribution for determining the probability of r successes is

$P\left(r\right){=}^n{C}_r\times q^{(n-r)}\times p^r$

Probabibility of getting exactly 4 heads = $P(r=4)$

$⟹P\left(4\right){=}^{10}{C}_4\times {\left(\frac{1}{2}\right)}^{(10-4)}\times {\left(\frac{1}{2}\right)}^4=\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}\times {\left(\frac{1}{2}\right)}^{10}=210\times {\left(\frac{1}{2}\right)}^{10}..............\left(i\right)$

Probabibility of getting exactly 6 heads = $P(r=6)$

$⟹P\left(6\right){=}^{10}{C}_6\times {\left(\frac{1}{2}\right)}^{(10-6)}\times {\left(\frac{1}{2}\right)}^6=\frac{10\times 9\times 8\times 7\times 6\times 5}{6\times 5\times 4\times 3\times 2\times 1}\times {\left(\frac{1}{2}\right)}^{10}=210\times {\left(\frac{1}{2}\right)}^{10}.....\left(ii\right)$

From equation (i) and (ii)

We can conclude that if 10 coins are thrown simultaneously, then the probability of exactly 4 heads appearing is equal to the probability of exactly 6 heads appearing.

And we can also observe that, ${}^{10}{C}_4{=}^{10}{C}_6⟹x=y$ or $n=x+y$ and as $p=q$ hence $P\left(H\right)=P\left(T\right)$ in a single toss.