Probability Ques 3
- Two numbers are selected randomly from the set $S=\{1,2,3,4,5,6\}$ without replacement one by one. The probability that minimum of the two numbers is less than $4$ , is
(2003, 1M)
(a) $1 / 15$
(b) $14 / 15$
(c) $1/5$
(d) $4 / 5$
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Answer:
Correct Answer: 3.(d)
Solution: (d) Here, two numbers are selected from $\{1,2,3,4,5,6\}$ $\Rightarrow n(S)=6 \times 5$ $\{$ as one by one without replacement $\} $
Favourable events $=$ the minimum of the two numbers is less than $4$. $n(E)=6 \times 4$ as for the minimum of the two is less than $4$ we can select one from $(1,2,3,4)$ and other from $(1,2,3,4,5,6)$
$\therefore \quad $ Required probability $=\frac{n(E)}{n(S)}=\frac{24}{30}=\frac{4}{5}$
BETA
