Question

Find the probability distribution of:
(i)Number of heads in two tosses of a coin
(ii)Number of tails in the simultaneous tosses of three coins
(iii)Number of heads in four tosses of a coin.

Answer 1

(i) When one coin is tossed twice, the sample space is ${HH, HT, TH, TT}$
Let X represent the number of heads.
$∴ X (HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0$

Therefore, X can take the value of 0, 1, or 2. It is known that,
P(HH)=P(HT)=P(TH)=P(TT)=$$\frac{1}{4}
P (X=0)=P(TT)=$\frac{1}{4}$
$P (X=1)=P(HT)+P(TH)$=$\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
P (X = 2) = P (HH)
Thus, the required probability distribution is as follows.

X	0	1	2
P(X)	$\frac{1}{4}$	$\frac{1}{2}$	$\frac{1}{4}$

(ii) When three coins are tossed simultaneously, the sample space is
{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
Let X represent the number of tails.
It can be seen that X can take the value of 0, 1, 2, or 3.
P (X=0)=P(HHH)=$\frac{1}{8}$
P (X=1)=P(HHT)+P(HTH)+P(THH) =$\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$
P (X = 2) = P (HTT) + P (THT) + P (TTH) =$\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$
P (X = 3) = P (TTT) =$\frac{1}{8}$
Thus, the probability distribution is as follows.

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

(iii) When a coin is tossed four times, the sample space is
S={HHHH,HHHT,HHTH,HHTT,HTHT,HTHH,HTTH,HTTT,THHH,THHT,THTH,THTT,TTHH,TTHT,TTTH,TTTT}
Let X be the random variable, which represents the number of heads.
It can be seen that X can take the value of 0, 1, 2, 3, or 4.
P (X = 0) = P (TTTT) =$\frac{1}{8}$
P (X = 1) = P (TTTH) + P (TTHT) + P (THTT) + P (HTTT)
=$\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}$
P (X = 2) = P (HHTT) + P (THHT) + P (TTHH) + P (HTTH) + P (HTHT) + P (THTH)
$=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{6}{16}=\frac{3}{8}$
P (X = 3) = P (HHHT) + P (HHTH) + P (HTHH) P (THHH)
=$\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}$
P (X = 4) = P (HHHH) =$\frac{1}{16}$
Thus, the probability distribution is as follows.

X	0	1	2	3	4
P(X)	$\frac{1}{16}$	$\frac{1}{4}$	$\frac{3}{8}$	$\frac{1}{4}$	$\frac{1}{16}$