Question

Assertion :If 8 coins are thrown simultaneously, then probability of occurence of exactly 2 tails is equal to the probability of occurence of exactly 6 tails. Reason: The binomial distribution probability of r success in n trial is calculated as $P(X=r)={}^n{\text{C}}_r{p}^r{q}^{n-r}$.

• 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 = 8

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

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

Probability of getting exactly 2 tails= $P(r=2)$

$⟹P\left(2\right){=}^8{C}_2\times {\left(\frac{1}{2}\right)}^{(8-2)}\times {\left(\frac{1}{2}\right)}^2=\frac{8\times 7}{2\times 1}{\left(\frac{1}{2}\right)}^8=28\times {\left(\frac{1}{2}\right)}^8..............\left(i\right)$

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

$⟹P\left(6\right){=}^8{C}_6\times {\left(\frac{1}{2}\right)}^{(8-6)}\times {\left(\frac{1}{2}\right)}^6=\frac{8\times 7\times 6\times 5\times 4\times 3}{6\times 5\times 4\times 3\times 2\times 1}\times {\left(\frac{1}{2}\right)}^8=28\times {\left(\frac{1}{2}\right)}^8.....\left(ii\right)$

From equation (i) and (ii)

We can conclude that if 8 coins are thrown simultaneously, then probability of occurence of exactly 2 tails is equal to the probability of occurence of exactly 6 tails.