Question

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.

Answer 1

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=$\frac{1}{4}$

∴$P(T)=\frac{1}{4} and P(H)=\frac{3}{4}$

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 (no tail) = P (H) × P (H) =$\frac{3}{4}X\frac{3}{4}=\frac{9}{16}$

P (X = 1) = P (one tail) = P (HT) + P (TH)

=$\frac{3}{4}.\frac{1}{4}+\frac{1}{4}.\frac{3}{4}$

=$\frac{3}{16}+\frac{3}{16}$

=$\frac{3}{8}$

P (X = 2) = P (two tails) = P (TT) =$\frac{1}{4}X\frac{1}{4}=\frac{1}{16}$

Therefore, the required probability distribution is as follows.

X	0	1	2
P(X)	$\frac{9}{16}$	$\frac{3}{8}$	$\frac{1}{16}$