Four fair coins are tossed simultaneously. The probability that at least one heads and at least one tails turn up is?

Four fair coins are tossed simultaneously. The probability that at least one heads and at least one tails turn up is? 

(A) $\dfrac{1}{16}$,    (B) $\dfrac{1}{8}$,    (C) $\dfrac{7}{8}$,    (D) $\dfrac{15}{16}$. 

(GATE 2002) 

Answer: (C) $\dfrac{7}{8}$ 

Explanation: 

Probability of getting head, $p = \dfrac{1}{2}$. 

Probability of getting tail, $q = \dfrac{1}{2}$ . 

Therefore, the probability of getting at least one head and one tail is 

$= {}^{4}\textrm{C}_{1} p^{1} q^{3} + {}^{4}\textrm{C}_{2} p^{2} q^{2} + {}^{4}\textrm{C}_{3} p^{3} q^{1}$, from binomial distribution 

$= {}^{4}\textrm{C}_{1}\left(\dfrac{1}{2}\right)^{1}\left(\dfrac{1}{2}\right)^{3} + {}^{4}\textrm{C}_{2}\left(\dfrac{1}{2}\right)^{2}\left(\dfrac{1}{2}\right)^{2} + {}^{4}\textrm{C}_{3}\left(\dfrac{1}{2}\right)^{3}\left(\dfrac{1}{2}\right)^{1}$ 

$= \dfrac{7}{8}$