Question

A bag contains four tickets with numbers 00,01,10,11. A ticket is drawn and replaced. In this way five tickets are drawn. Find the probability that the sum of the numbers on the ticket drawn is 23.

• A. $\frac{100}{4^5}.$
• B. $\frac{90}{4^5}.$
• C. $\frac{10}{4^5}.$
• D. $\frac{60}{4^5}.$

Answer 1

Answer: (A)

$\frac{100}{4^5}.$

The total ways of drawing 5 tickets with replacement out of 4 tickets $n=4^5.$
Now the following are the favourable cases of sum being 23 of the 5 tickets drawn :
1) $00,00,01,11,11$
These numbers can come in different order .So, number of ways of such arrangement is $\frac{5!}{2!2!1!}$
2) $00,01,01,10,11$
These numbers can come in different order .So, number of ways of such arrangement is $\frac{5!}{2!1!1!1!}$
3) $01,01,01,10,10$
These numbers can come in different order .So, number of ways of this arrangement is $\frac{5!}{3!2!}$
So, the total number of favourable cases are $\frac{5!}{2!2!1!}+\frac{5!}{2!1!1!1!}+\frac{5!}{3!2!}$
Hence the required probability $=\frac{5!}{2!2!1!}{\left(\frac{1}{4}\right)}^5+\frac{5!}{2!1!1!1!}{\left(\frac{1}{4}\right)}^5+\frac{5!}{3!2!}{\left(\frac{1}{4}\right)}^5$
$=5!{\left(\frac{1}{4}\right)}^5[\frac{1}{4}+\frac{1}{2}+\frac{1}{12}]$
$=\frac{1}{4^5}.120\left[\frac{10}{12}\right]=\frac{100}{4^5}.$