1.

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

Answer»

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) = \(\frac {n(B)}{n(S)} = \frac {4} {8}\)

Now, A ∩ B = {HHH} 

∴ n(A ∩ B) = 1

 ∴ P(A ∩ B) = \(\frac {n(A∩B)}{n(S)} = \frac {1} {8}\)

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

P(A/B) = \(\frac {P(A∩B)}{P(B)} \)

= \(\frac {\frac{1}{8}}{\frac{4}{8}}\)

= 1/4.


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