Question

When two dice are rolled, let $x$ be the probability of getting a sum of the numbers appear on the dice is at most $7$ . Let $y$ be the probability of getting a sum $7$ at least once when a pair of dice are rolled $n$ times. In order to have $y>x$. the minimum $n$ is

• A. 3
• B. 6
• C. 5
• D. 4

Answer 1

Answer: (C)

We have
$x=P($ getting sum at most 7$)$
$=P($ sum $=2)+P($ sum $=3)+P($ sum $=4)+P($ sum $=5)+P($ sum $=6)+P($ sum $=7)$
$=\frac{1}{36}+\frac{2}{36}+\frac{3}{36}+\frac{4}{36}+\frac{5}{36}+\frac{6}{36}=\frac{21}{36}$
And $y=P($ getting sum $=7$ atleast once when pair of dice rolled $n$ times $)$
$=1-P($ getting sum $=7$ zero times $)$
$=1-{ }^{n} C_{0}\left(\frac{6}{36}\right)^{0}\left(\frac{30}{36}\right)^{n}=1-\left(\frac{30}{36}\right)^{n}$
Now $y>x$
$1-\left(\frac{5}{6}\right)^{n}>\frac{7}{12}$
$1-\frac{7}{12}>\left(\frac{5}{6}\right)^{n}$
$\left(\frac{5}{6}\right)^{n}<\left(\frac{5}{12}\right)$
$ \Rightarrow \left(\frac{5}{6}\right)^{n}<\frac{5}{6} \times \frac{1}{2}$
$\left(\frac{5}{6}\right)^{n-1}<\frac{1}{2}$
$\therefore $ Minimum value of $n$ is $5$