Class 12

Math

Algebra

Probability I

A coin is tossed twice. If the outcome is at most one tail, what is the probability that both head and tail have appeared?

Then the sample space $S$ will be ${HH,HT,TH,TT}$

Given that the outcome is at most one tail

So favourable outcomes will be ${HH,TH,HT}$

Let $A$ be the event of getting at most one tail

$⟹n(A)=3,n(S)=4$

Probability of $A$ to happen is $P(A)=n(S)n(A) =43 $

Let $B$ be the event that both head and tail had appeared

then the outcomes of $B$ is ${HT,TH}$

$⟹n(B)=2$

Probability of $B$ to happen is $P(B)=n(S)n(B) =42 =21 $

Favourable outcomes of $A∩B$ is ${HT,TH}$

$⟹A∩B=B$

$⟹P(A∩B)=P(B)=21 $

Required probability is the probability of $B$ given that $A$ has happen $..i.e.,P(B/A)$

As we know that

$P(B/A)=P(A)P(A∩B) =43 21 =32 $

So Probability that both head and tail have appeared if the the outcome is at most one tail is $32 $