Assertion (A): If 8 coins are thrown simultaneously, then probability of occurrence of exactly 2 tails is equal to the probability of occurrence of exactly 6 tails. Reason (R): For binomial distribution, probability of r success in n trial is calculated as P(X = r) = nCr pr qn-r.

Question

Assertion (A): If 8 coins are thrown simultaneously, then probability of occurrence of exactly 2 tails is equal to the probability of occurrence of exactly 6 tails. Reason (R): For binomial distribution, probability of r success in n trial is calculated as P(X = r) = nCr pr qn-r. (A) Both (A) (R) are individually true (R) is correct explanation of (A). (B) Both (A) (R) are individually true but (R) is not the correct (proper) explanation of (A). (C) (A) is true but (R) is false. (D) (A) is false but (R) is true.

Tardigrade Admin · Accepted Answer

P(T) = P(H) = (1/2) (for one trial ), n = 8 ∴ P(X = r) = 8Cr((1/2))r((1/2))8-r = 8C2((1/2))2((1/2))6 = 8C6((1/2))6((1/2))2.

Both (A) & (R) are individually true & (R) is correct explanation of (A).

Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).

(A) is true but (R) is false.

(A) is false but (R) is true.

$P (T) = P (H) = \frac{1}{2}$ (for one trial ), $n = 8$
$∴ P (X = r) =^{8} C_{r} (\frac{1}{2})^{r} (\frac{1}{2})^{8 - r}$
$=^{8} C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{6} =^{8} C_{6} (\frac{1}{2})^{6} (\frac{1}{2})^{2}$ .

Q. Assertion (A) : If 8 coins are thrown simultaneously, then probability of occurrence of exactly 2 tails is equal to the probability of occurrence of exactly 6 tails. Reason (R) : For binomial distribution, probability of r success in n trial is calculated as P(X=r)=nCr​prqn−r.

Both (A) & (R) are individually true & (R) is correct explanation of (A).

Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).

(A) is true but (R) is false.

(A) is false but (R) is true.

P(T)=P(H)=21​ (for one trial ), n=8 ∴P(X=r)=8Cr​(21​)r(21​)8−r =8C2​(21​)2(21​)6=8C6​(21​)6(21​)2.

$P (T) = P (H) = \frac{1}{2}$ (for one trial ), $n = 8$
$∴ P (X = r) =^{8} C_{r} (\frac{1}{2})^{r} (\frac{1}{2})^{8 - r}$
$=^{8} C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{6} =^{8} C_{6} (\frac{1}{2})^{6} (\frac{1}{2})^{2}$ .