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,

$$\begin{gathered} p=\frac{13}{52}=\frac{1}{4} \\ \mathrm{and}q=1-q=1-\frac{1}{4}=\frac{3}{4} \end{gathered}$$

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) < $\frac{1}{4}$
$$\begin{gathered} {}^{n}C_0{\left(\frac{1}{4}\right)}^0{\left(\frac{3}{4}\right)}^{n-0}<\frac{1}{4} \\ {\left(\frac{3}{4}\right)}^n<\frac{1}{4} \\ \mathrm{Put}n=1,{\left(\frac{3}{4}\right)}^1\mathrm{not}\mathrm{less}\mathrm{than}\frac{1}{4} \\ n=2,{\left(\frac{3}{4}\right)}^2\mathrm{not}\mathrm{less}\mathrm{than}\frac{1}{4} \\ n=3,{\left(\frac{3}{4}\right)}^3\mathrm{not}\mathrm{less}\mathrm{than}\frac{1}{4} \\ \mathrm{So},\mathrm{smallest}\mathrm{value}\mathrm{of}n=3 \end{gathered}$$
Therefore card must be drawn three times.

(ii) Given the probability of drawing a heart > $\frac{3}{4}$
1 - P(X=0) > $\frac{3}{4}$
$$\begin{gathered} 1-{}^{n}C_0{\left(\frac{1}{4}\right)}^0{\left(\frac{3}{4}\right)}^{n-0}>\frac{3}{4} \\ 1-{\left(\frac{3}{4}\right)}^n>\frac{3}{4} \\ 1-\frac{3}{4}>{\left(\frac{3}{4}\right)}^n \\ \frac{1}{4}>{\left(\frac{3}{4}\right)}^n \end{gathered}$$

$$\begin{gathered} Forn=1,{\left(\frac{3}{4}\right)}^{1}notlessthan\frac{1}{4} \\ n=2,{\left(\frac{3}{4}\right)}^{2}notlessthan\frac{1}{4} \\ n=3,{\left(\frac{3}{4}\right)}^{3}notlessthan\frac{1}{4} \\ n=4,{\left(\frac{3}{4}\right)}^{4}notlessthan\frac{1}{4} \\ n=5,{\left(\frac{3}{4}\right)}^{5}notlessthan\frac{1}{4} \end{gathered}$$
So, card must be drawn 5 times.