Question

A coin is biased so that the head is 3 times as likely to occur as a tail. If the coin is tossed twice, find the probability distribution of the number of tails.

Answer 1

If a coin is tossed, the probability of getting head is $\dfrac{1}{2}$ and probability of getting tail is $\dfrac{1}{2}$. In this question, it is given that the coin is tossed twice and a coin is biased so that the head is $3$times as likely to occur as a tail. Therefore, the probability of getting head is $\dfrac{3}{4}$ and the probability of getting tail is $\dfrac{1}{4}$. We will find the probability distribution of the number of tails by using the outcomes of the coin.

Complete step-by-step solution:
Given that the coin is tossed twice and also given that the head is $3$ times as likely to occur as a tail. i.e 4 times a coin is tossed the probability of getting a head is $3$ times, and the probability of getting a tail is$1$time.
Therefore, the probability of getting head P(H) = $\dfrac{3}{4}$ ………..$(1)$
And, the probability of getting tail P(T) = $\dfrac{1}{4}$ ……………………$(2)$
Since the coin is tossed twice, we were asked to find the probability distribution of the number of tails.
The coin is tossed twice. Therefore the outcomes will be

Two heads (or)
In the first toss, there will be a head and in the second toss there will be a tail (or) In the first toss there will be a tail and in the second toss there will be a head (or)
Two tails.
Let X be the event of getting the number of tails. Since the coin is tossed twice, we can get no tails or one-tail or two tails. Therefore, X can take the values $0,1$ and $2$.
When X=0, the outcome is {HH}
Therefore, the probability of X, P(X) = P(H)$\times$ P(H)
From equation$(1)$, We know that P(H) = $\dfrac{3}{4}$, substituting the value of P(H) in above we get
P(X) = $\dfrac{3}{4}$ $\times$ $\dfrac{3}{4}$= $\dfrac{9}{{16}}$
When X=1, the outcome is {HT, TH}
Therefore, the probability of X, P(X) = P(H)$\times$ P(T) + P(T)$\times$ P(H)
From equation $(1)$and$(2)$, We know that, P(H) = $\dfrac{3}{4}$and P(T) = $\dfrac{1}{4}$, substituting the value of P(H)and P(T) in above we get
P(X) = $\dfrac{3}{4}$ $\times$ $\dfrac{1}{4}$+$\dfrac{1}{4}$ $\times$ $\dfrac{3}{4}$
= $\dfrac{3}{{16}}$ + $\dfrac{3}{{16}}$
= $\dfrac{6}{{16}}$
When X=2, the outcome is {TT}
Therefore, the probability of X, P(X) = P(T) $\times$P(T)
From equation$(2)$, We know that P(T) = $\dfrac{1}{4}$, substituting the value of P(T) in above we get
P(X) = $\dfrac{1}{4}$ $\times$ $\dfrac{1}{4}$
= $\dfrac{1}{{16}}$
Therefore the probability distribution is

X	0	1	2
P(X)	$\dfrac{9}{{16}}$	$\dfrac{6}{{16}}$	$\dfrac{1}{{16}}$

Note: This problem varies according to the number of times the coin is tossed and the possibilities of outcomes that are given in the question. In this question, we were asked to find the probability distribution of the number of tails. If we are asked to find heads the probability will change and so the probability distribution.