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14.Probability
easy
Seven chits are numbered $1$ to $7$. Three are drawn one by one with replacement. The probability that the least number on any selected chit is $5$, is
A
$1 - {\left( {\frac{2}{7}} \right)^4}$
B
$4\,{\left( {\frac{2}{7}} \right)^4}$
C
${\left( {\frac{3}{7}} \right)^3}$
D
None of these
Solution
(c) $P(5\,{\rm{or}}\,6\,{\rm{or}}7)$ in one draw $ = \frac{3}{7}$
$\therefore $ Probability that in each of $3$ draws, the chits bear $5$ or $6$ or $7$
$= {\left( {\frac{3}{7}} \right)^3}.$
Standard 11
Mathematics
