Probability — Class 12 Maths Solution

.: (i) The sample space will be $= \{ HH,HT,TH,TT\} .$ Let $X$ denote the random
variable which represents the number of heads.

$\therefore$ $X$ can assume values 0, 1 and 2.
Hence, the probability distribution is

$X$ 0 1 2

$P(X)$ $P(TT) = \cfrac{1}{4}$ $P(TH,HT) = \cfrac{1}{2}$ $P(HH) = \cfrac{1}{4}$

(ii) The sample space will be $= \{ HHH,HHT,HTH,THH,TTH,THT,HTT,TTT\}$

Let $X$ denote the random variable which

represents the number oftails. $X$ can assume values 0, 1, 2 and 3.

$\therefore$ $P(X = 0) = P(HHH) = \cfrac{1}{8}$

$P(X = 1) = P(HHT,HTH,THH) = \cfrac{3}{8}$

$P(X = 2) = P(TTH,THT,HTT) = \cfrac{3}{8}$

and $P(X = 3) = P(TTT) = \cfrac{1}{8}$

Hence, the probability distribution is :

(iii) Here, the sample space will be $$\begin{array}{l} = HHHH,{\rm{ }}HHHT,{\rm{ }}HHIH,{\rm{ }}HTHH,\\THHH,{\rm{ }}HHTT,{\rm{ }}HTHT,{\rm{ }}HTTH,\\THH{\rm{ }}T,{\rm{ }}THT{\rm{ }}H,{\rm{ }}TTHH,{\rm{ }}HTTT,\\THTT,TIHT,TTTH,TTTT\end{array}$$

Let $X$ denote random variable which represents the number of heads. $X$ can assume values 0, 1, 2, 3 and 4

$\therefore$ $P(X = 0) = P(TTTT) = \cfrac{1}{{16}}$

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

$P(X = 2) = (\{ HHTT,HTHT,HTTH,THHT,THTH,TTHH\} = \cfrac{6}{{16}} = \cfrac{3}{8}$

$P(X = 3) = P(\{ HHHT,HHTH,HTHH,THHH\} = \cfrac{4}{{16}} = \cfrac{1}{4}$

$P(X = 4) = P(HHHH) = \cfrac{1}{{16}}$

Hence, the probability distribution is :