Question

Two different coins are tossed simultaneously. Find the probability of getting:-
(i) at least one head
(ii) at least one tail
(iii) at least two head
(iv) at lest two tail
(v) at most one head
(vi) at most one tail
(vii) at most two head
(viii) at most two tail.

Answer 1

Dear Student,

$$\begin{gathered} Whentwocoinsaretossedthentheoutcomewillbe:TT,HT,TH,HH. \\ Hencetotalnumberofoutcomeis4. \\ Probability=\frac{Numberoffavourableoutcomes}{Totalnumberofoutcomes} \\ \left(i\right)atleastonehead. \\ Favourableoutcomes=\{HT,TH,HH\}=3 \\ Probabilityofgettingatleastonehead=\frac{3}{4} \\ \left(ii\right)atleastonetail. \\ Favourableoutcomes=\{HT,TH,TT\}=3 \\ Probabilityofgettingatleastonetail=\frac{3}{4} \\ \left(iii\right)atleast2heads \\ Favourableoutcomes=\left\{HH\right\}=1 \\ Probabilityofgettingatleasttwoheads=\frac{1}{4} \\ \left(iv\right)atleast2tails \\ Favourableoutcomes=\left\{TT\right\}=1 \\ Probabilityofgettingatleasttwotails=\frac{1}{4} \\ \left(v\right)atmostonehead \\ Favourableoutcomes=\{HT,TH,TT\}=3 \\ Probabilityofgettingatmostonehead=\frac{3}{4} \\ \left(vi\right)atmostonetail \\ Favourableoutcomes=\{HT,TH,HH\}=3 \\ Probabilityofgettingatmostonetail=\frac{3}{4} \\ \left(vii\right)atmosttwoheads \\ Favourableoutcomes=\{HT,TH,TT,HH\}=4 \\ Probabilityofgettingatmosttwoheads=\frac{4}{4}=1 \\ \left(viii\right)atmosttwotails \\ Favourableoutcomes=\{HT,TH,TT,HH\}=4 \\ Probabilityofgettingatmosttwotails=\frac{4}{4}=1 \end{gathered}$$

Regards