There are three indentical balls marked $A, B$ and $C$ randomly shuffled in a bag. Consider the experiment that balls are drawn out from the bag until the ball marked '$C$' is drawn. We have to find the probability that the first ball drawn will be marked '$C$'.

I tried doing the problem in $2$ ways-

1) The possible outcomes of the experiment are $C, AC, BC, ABC, BAC$ out of which only the first is favourable. Hence probability $=1/5$.

2) There are $3!=6$ ways of drawing the $3$ balls out of which $2$ ways ($CAB$ and $CBA$) are favourable. Hence probability $=1/3$.

The first method seems more true to the experiment while the second one seems more correct intuitively to me. Which one is correct?

Also, if the experiment was performed with three cards marked $A, B$ and $C$ instead, what would be the correct probability? In this case, the second method seems much more logical to me as, unlike in the first experiment wherein the balls are randomly drawn, in this case once the cards are shuffled, their order becomes fixed.

  • $\begingroup$ The likelihoods in part 1 are not equal. P(C)= 1/3 but P(AC) = (1/3)(1/2) $\endgroup$ – Remy Nov 13 '17 at 6:17
  • $\begingroup$ I still don't get why I can't directly use P = No. of favourable outcomes/Total no. of outcomes though. Isn't that formula always valid? $\endgroup$ – User Nov 13 '17 at 6:28
  • 1
    $\begingroup$ That approach will not always work. For example, a sniper shoots at you and he hits you with probability $.8$. Then the outcomes are he hits you or he does not. Does that mean the probability is $\frac{1}{2}$? $\endgroup$ – Remy Nov 13 '17 at 6:35
  • 1
    $\begingroup$ @InternetGuy, you might want to look at math.meta.stackexchange.com/q/27347/18398 $\endgroup$ – Joel Reyes Noche Nov 13 '17 at 7:09

You have the correct possible outcomes, and to me, that seems like the easier approach.

$$C, AC, BC, ABC, BAC$$

The probability of these, respectively, are

$$\frac{1}{3}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6},\frac{1}{6}$$

$$P(C\text{ is first ball chosen})=\frac{\frac{1}{3}}{\frac{1}{3}+ \frac{1}{6}+ \frac{1}{6}+ \frac{1}{6}+\frac{1}{6}} = \frac{1}{3}$$

This is the same as the probability of selecting $C$ first, without the constraint that we select balls until we draw $C$.


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.