Question

A card is drawn and replaced in an ordinary pack of 52 cards. How many times must a card be drawn so that (i) there is at least an even chance of drawing a heart (ii) the probability of drawing a heart is greater than 3/4?

Answer 1

(i) Let p denote the probability of drawing a heart from a deck of 52 cards. So,

\[p = \frac{13}{52} = \frac{1}{4}\]
\[\text{ and }  q = 1 - q = 1 - \frac{1}{4} = \frac{3}{4}\] 

Let the card be drawn n times. So, binomial distribution is given by:  \[P(X = r) = ^{n}{}{C}_r p^r q^{n - r}\]

Let X denote the number of hearts drawn from a pack of 52 cards.
We have to find the smallest value of n for which P(X=0) is less than \[\frac{1}{4}\] P(X=0) < \[\frac{1}{4}\] 

\[^{n}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{n - 0} < \frac{1}{4}\]

\[ \left( \frac{3}{4} \right)^n < \frac{1}{4}\]
\[\text{ Put n}  = 1, \left( \frac{3}{4} \right)^1 \text{ not less than } \frac{1}{4}\]
\[ n = 2, \left( \frac{3}{4} \right)^2 \text{ not less than}  \frac{1}{4}\]
\[ n = 3, \left( \frac{3}{4} \right)^3 \text{ not less than }\frac{1}{4}\]
\[\text{ So, smallest value of n } = 3\]

Therefore card must be drawn three times.
(ii) Given the probability of drawing a heart > \[\frac{3}{4}\]

1 - P(X=0) > \[\frac{3}{4}\]
\[1 -^{n}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{n - 0} > \frac{3}{4}\]
\[1 - \left( \frac{3}{4} \right)^n > \frac{3}{4}\]
\[1 - \frac{3}{4} > \left( \frac{3}{4} \right)^n \]
\[\frac{1}{4} > \left( \frac{3}{4} \right)^n\]

\[\text{ For n}  = 1, \left( \frac{3}{4} \right)^1 \text{ not less than } \frac{1}{4}\]
\[ n = 2, \left( \frac{3}{4} \right)^2 \text{ not less than } \frac{1}{4}\]
\[ n = 3, \left( \frac{3}{4} \right)^3 \text{ not less than } \frac{1}{4}\]
\[ n = 4, \left( \frac{3}{4} \right)^4 \text{ not less than } \frac{1}{4}\]
\[ n = 5, \left( \frac{3}{4} \right)^5 \text{ not less than } \frac{1}{4}\]

So, card must be drawn 5 times.