An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is :

Question

An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is : (A) (255/256) (B) (127/128) (C) (63/64) (D) (1/2)

Tardigrade Admin · Accepted Answer

1 - ∫ P(All Head) + P(AllTail) 1 - (1/28)+(1/28) = 1-(1/27) = 1-(1/128) = (127/128)

Q. An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is :

$\frac{255}{256}$

$\frac{127}{128}$

$\frac{63}{64}$

$\frac{1}{2}$

1 - $\int$ P(All Head) + P(AllTail)
$1 - {\frac{1}{2 ^{8}} + \frac{1}{2 ^{8}}}$
$= 1 - \frac{1}{2 ^{7}}$
$= 1 - \frac{1}{128} = \frac{127}{128}$

Q. An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is :

256255​

128127​

6463​

21​

1 - ∫ P(All Head) + P(AllTail) 1−{281​+281​} =1−271​ =1−1281​=128127​

$\frac{255}{256}$

$\frac{127}{128}$

$\frac{63}{64}$

$\frac{1}{2}$

1 - $\int$ P(All Head) + P(AllTail)
$1 - {\frac{1}{2 ^{8}} + \frac{1}{2 ^{8}}}$
$= 1 - \frac{1}{2 ^{7}}$
$= 1 - \frac{1}{128} = \frac{127}{128}$