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In tossing three coins at a time, what is the probability of getting at most one head?
Options:
Solution:
2850 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}$
Possible samples are as follows $\{$ HHH, HTH, HHT, THH, TTH, THT, HTT, TTT\}
Let $A$ be the event of getting one head. Let $\mathrm{B}$ be the event of getting no head. Favourable outcome for
$A=\{T T H, T H T, H T T\}$
Favourable outcome for
$$
B=\{T T T\}
$$
Total no. of outcomes $=8$
$$
\therefore \quad P(A)=\frac{3}{8}, P(B)=\frac{1}{8}
$$
$\therefore$ Required probability $=$ Probability of getting one head
+ Probability of getting no head
$$
=P(A)+P(B)=\frac{3}{8}+\frac{1}{8}=\frac{4}{8}=\frac{1}{2}
$$
Let $A$ be the event of getting one head. Let $\mathrm{B}$ be the event of getting no head. Favourable outcome for
$A=\{T T H, T H T, H T T\}$
Favourable outcome for
$$
B=\{T T T\}
$$
Total no. of outcomes $=8$
$$
\therefore \quad P(A)=\frac{3}{8}, P(B)=\frac{1}{8}
$$
$\therefore$ Required probability $=$ Probability of getting one head
+ Probability of getting no head
$$
=P(A)+P(B)=\frac{3}{8}+\frac{1}{8}=\frac{4}{8}=\frac{1}{2}
$$
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