2
$\begingroup$

The Question I encountered was:

"A box contains one red and three identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment."

If we denote Red Ball with $R$ and white ball with $W$, the sample space given in the answer is {$RW$$,WR$$,WW$}. Now here is my doubt, suppose I want to know the probability of drawing one red ball then from this sample space it is $2/3$ while the probability of drawing two white balls is $1/3$ even though the number of white balls are more. If I solve the problem of drawing one red ball in the following way, I get the probability as $1/2$. Where am I wrong?

Required Probability $= RW$ or $WR$ $=$ $\frac{1}{4}\times$ $1 + \frac{3}{4}\times\frac{1}{3} = \frac{1}{2}$

$\endgroup$
4
$\begingroup$

Each event in a sample space doesn't have to have equal probability. This is true for many of the famous distributions like Binomial, Poisson, etc.

The probability of getting the red is $1/2$. Your approach looks good and is leveraging the fact that RW and WR are mutually exclusive and then applying the multiplication rule.

$$ P(\text{draw the red ball}) = P(RW \cup WR) = P(RW) + P(WR) = P(R)P(W|R) + P(W)P(R|W) = 1/4(1) + 3/4(1/3) = 1/2$$

You could also see this from the pmf of a hypergeometric RV.

Let $X =$ # of red balls you get from 2 draws.

$$ P(X=1) = \frac{{1 \choose 1}{3 \choose 1}}{{4 \choose 2}} = 1/2$$

If you want each event to have equal probability in your sample space it could help to label the balls. So there are $12$ events in the sample space $R_1 W_1, R_1 W_2, R_1 W_3, ...$ and then you can treat these events as equiprobable and can count the events. Then the probability of one red will be $6/12 = 1/2$

$\endgroup$
4
$\begingroup$

The probability of drawing a red ball is not 2/3, because all the points in the sample space do have equal probabilities. \begin{equation} \mathbb{P}\{RW\} = \frac{1}{4} \hspace{0.5cm} \mathbb{P}\{WR\} = \frac{3}{4}.\frac{1}{3} = \frac{1}{4} \hspace{0.5cm} \mathbb{P}\{WW\} = \frac{3}{4}\frac{2}{3} = \frac{1}{2}. \end{equation} Thus, the required probability is $\mathbb{P}\{RW\} +\mathbb{P}\{WR\} = \frac{1}{2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.