Question

Three fair coins are tossed. What is the probability of getting three heads given that at least two coins show heads?

Answer 1

When three fair coins are tossed, the sample space is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

∴ n(S) = 8

Let event A: Getting three heads.

∴ A = {HHH}

Let event B: Getting at least two heads.

∴ B = {HHT, HTH, THH, HHH}

∴ n(B) = 4

∴ P(B) = `("n"("B"))/("n"("S")) = 4/8`

Now, A ∩ B = {HHH}

∴ n(A ∩ B) = 1

∴ P(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S")) = 1/8`

∴ Probability of getting three heads, given that at least two coins show heads, is given by

`"P"("A"/"B") = ("P"("A" ∩ "B"))/("P"("B")`

= `(1/8)/(4/8)`

= `1/4`