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)	\(\frac{1}{16}\)	\(\frac{1}{4}\)	\(\frac{3}{8}\)	\(\frac{1}{4}\)	\(\frac{1}{16}\)