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

$\begin{array}{1 1}(A)\;\large\frac{7}{5}\\(B)\;\large\frac{7}{8}\\(C)\;\large\frac{7}{2}\\(D)\;\text{None of these}\end{array} $

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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 heads [max 2 heads can consider 0 head,1 head,2head]
Let E be the event of getting atmost 2 heads
$\therefore n(E)=$2 heads + 1 head + 0 head
$\Rightarrow 3+3+1$
$\Rightarrow 7$
Step 3:
$\therefore$ Probability of getting atmost 2 heads=$\large\frac{n(E)}{n(S)}$
$\Rightarrow \large\frac{7}{8}$
Hence (B) is the correct answer.
answered Jul 1, 2014 by sreemathi.v
 
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