Question

Eight coins are tossed once, find the probability of getting exactly two tails

Answer 1

Eight coins are tossed simultaneously one time = one coin is tossed eight times.

Let S be the sample space.

S = {H, T} × {H, T} × ………….. × {H, T} 8 times

Let A be the event of getting exactly two heads,

B be the event of getting atleast two tails and

C be the event of getting atmost two tails.

When eight coins are tossed, the number of elements in the sample space

n(S) = 2⁸ = 256

n(A) = 8C₂

= `(8 xx 7)/(1 xx 2)`

= 28

n(B) = 8C₂ + 8C₃ + 8C₄ + 8C₅ + 8C₆ + 8C₇ + 8C₈

= n(S) – (8C₈ + 8C₁)

= n(S) – {n(Event of getting all heads) + n(Event of getting one head)}

= n(S) – (1 + 8)

= 256 – 9

= 247

n(C) = 8C₀ + 8 C₁ + 8 C₂

= `1 + 8 + (8 xx 7)/(1 x 2)`

= 1 + 8 + 28

= 37

P{getting exactly two tails) =

P(A) = `("n"("A"))/("n"("S"))`

P(A) = `28/256`

= `7/64`