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Three coins are tossed once.Find the probability of getting atmost 2 tails

$\begin{array}{1 1}(A)\;\large\frac{3}{8}\\(B)\;\large\frac{7}{8}\\(C)\;\large\frac{5}{8}\\(D)\;\text{None of these}\end{array} $

1 Answer

Step 1:
If three coins are tossed,then the possible outcomes =$2^3=8$
$\therefore$ The sample space $S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$
Step 2:
Atmost 2 tails [max 2 tails,can consider 0 tail,1 tail & 2 tail]
Let E be the event of getting atmost 2 tail
$\therefore$ n(E)=2 tails + 1 tail + 0 tail
$\Rightarrow 3+3+1$
$\Rightarrow 7$
Step 3:
Probability of getting atmost 2 tails =$\large\frac{n(E)}{n(S)}=\frac{7}{8}$
Hence (B) is the correct answer.
answered Jul 1, 2014 by sreemathi.v
 
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