Question

Find the probability distribution of

number of heads in two tosses of a coin

number of tails in the simultaneous tosses of three coins

number of heads in four tosses of a coin.

Answer 1

When one coin is tossed twice, the sample space is

S={HH,HT,TH,TT}.

Let X denotes, the number of heads in any outcome in S,

X(HH)=2, X(HT) =1, X(TH)= 1 and 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)= $\frac{1}{4}$

$\therefore$ P(X=0) =P (tail occurs on both tosses) $=P\left(TT\right)=\frac{1}{4}$

P(X=1)=P (one head and one tail occurs) =$P\left(TH,HT\right)=\frac{2}{4}=\frac{1}{2}$

and P(X=2) =P(head occurs on both tosses) =$P\left(HH\right)=\frac{1}{4}$

Thus, the required probability distribution is as follows

$\begin{array}{cccc}X& 0& 1& 2\\ P\left(X\right)& \frac{1}{4}& \frac{1}{2}& \frac{1}{4}\end{array}$

When three coins are tossed thrice, the sample space is

S= {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}

Which contains eight equally likely sample points.

Let X represent the number of tails.

Then, X can take values 0,1,2, and 3.

P(X=0) = P(no tail) =P({HHH})=$\frac{1}{8}$,

P(X=1) =P (one tail and two heads show up)

= P ({HHT,HTH,THH})=$\frac{3}{8},$

P(X=2) = P(two tails and one head show up)

= P ({HTT,THT,TTH})=$\frac{3}{8}$

and P(X=3) =P (three tails show up) =P({TTT}) =$\frac{1}{8}$.

Thus , the probability distribution is as follows

$\begin{array}{ccccc}X& 0& 1& 2& 3\\ P\left(X\right)& \frac{1}{8}& \frac{3}{8}& \frac{3}{8}& \frac{1}{8}\end{array}$

When a coin is tossed four times, the sample space is

$\left\{\begin{array}{cccccccc}HHHH& HHHT& HHTH& HTHT& HTTH& HTTT& THHH& HTHH\\ THHT& THTH& HHTT& TTHH& TTHT& TTTH& THTT& TTTT\end{array}\right\}$

which contains 16equally likely sample points.

Let X be the random variable, which represent the number of heads. It can be seen that X can take the value of 1,2,3, or 4.

P(X=0)=P (no head shows up) = P{TTTT} =$\frac{1}{16}$,

P(X=1)= P (one head and three tails show up)

= P(HTTT,THTT,TTHT,TTTH)=$\frac{4}{16}=\frac{1}{4},$

P(X=2) =P (two heads and two tails show up)

= P({HHTT,HTHT,HTTH,THHT,THTH,TTHH}) =$\frac{6}{16}=\frac{3}{8}$,

P(X=3) =P (three heads and one tail show up)

= P({HHHT,HHTH,HTHH,THHH})= $\frac{4}{16}=\frac{1}{4}$

and P(X=4)=P (four heads show up) = P({HHHH}) =$\frac{1}{16}$

Thus, the probability distribution is as follows:

$\begin{array}{cccccc}X& 0& 1& 2& 3& 4\\ P\left(X\right)& \frac{1}{16}& \frac{1}{4}& \frac{3}{8}& \frac{1}{4}& \frac{1}{16}\end{array}$