Question

Three coins are tossed together. Find the probability of getting:
(i) exactly two heads (ii) at most two heads (iii) at least one head and one tail (iv) no tails

Answer 1

When three coins are tossed together, the total number of outcomes $=8$

i.e., $(HHH,HHT,HTH,THH,TTH,THT,HTT,TTT)$

Solution (i):

Let $E$ be the event of getting exactly two heads

Therefore, no. of favorable events, $n\left(E\right)=3(i.e.,HHT,HTH,THH)$

We know that, $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{3}{8}$

Solution (ii):

Let $F$ be the event of getting atmost two heads

Therefore, no. of favorable events, $n\left(E\right)=7(i.e.,HHT,HTH,TTT,THH,TTH,THT,HTT)$

We know that, $P\left(F\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{7}{8}$

Solution (iii):

Let $H$ be the event of getting at least one head and one tail

Therefore, no. of favorable events, $n\left(H\right)=6(i.e.,HHT,HTH,THH,TTH,THT,HTT)$

We know that, $P\left(H\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{6}{8}$$=\frac{3}{4}$

Solution (iv):

Let $I$ be the event of getting no tails

Therefore, no. of favorable events, $n\left(I\right)=1(i.e.,HHH)$

We know that, $P\left(H\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{1}{8}$