A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.
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Solution
Let the probability of getting a tail in the biased coin be x.
∴P(T)=x
⇒P(H)=3x
For a biased coin, P(T)+P(H)=1
⇒x+3x=1
⇒4x=1
⇒x=14
∴P(T)=14 and P(H)=34
When the coin is tossed twice, the sample space is {HH,TT,HT,TH}.
Let X be the random variable representing the number of tails. ∴P(X=0)=P(notail)=P(H)×P(H)=34×34=916
P(X=1)=P(onetail)=P(HT)+P(TH)
=34⋅14+14⋅34
=316+316
=38
P(X=2)=P(twotails)=P(TT)=14×14=116
Therefore, the required probability distribution is as follows.