[Math] Black and White Marbles Probability Question

probability

I found this question in a SAT Math 2 test:

A jar contains two black marbles and one white marble. If a marble is selected from the jar and removed, and then a second marble is selected form the jar and removed, what is the probability the second marble is white?

If a person takes one marble out of the jar, then the 1/3 of the jar's marbles would be removed. This would leave the person with two combinations: 1 black marble and 1 white marble or 2 black marbles. The former is more likely according to the laws of probability, so the 2nd time the marble is removed, the probability would be 1/2 for the marble being white.

I am confused on whether I am supposed to add 1/3 and 1/2 to determine the overall probability of the second picked marble being white or multiple the two numbers. Also just in general, when would we add and multiply probability values?

Best Answer

There are a number of ways of approaching this problem, but the easiest solution is to realize that it doesn't matter what order you took the marbles out in. Take out a marble. Call it the "second" marble. Trivially then, the answer is $\frac{1}{3}$ since there is one white.

The other approach is more tedious. Consider what happens for the first marble. Either it is black or it is white.

The probability that it was black was $\frac{2}{3}$ and the probability that it was white was $\frac{1}{3}$

  • If the first was a black, there now remain one black and one white marble. The chance that the next marble drawn is white is now $\frac{1}{2}$. By multiplication principle then this occurs with probability $\frac{2}{3}\cdot \frac{1}{2} = \frac{1}{3}$

  • If the first marble was white, the second cannot be white (as there are no whites left in the urn/bag). Thus to arrive at this leaf of the probability tree by multiplication principle it occurs with probability $\frac{1}{3}\cdot 0 = 0$

The total probability then is the sum of these: $\frac{1}{3} + 0 = \frac{1}{3}$

In general, you will only add probabilities together if they are disjoint/mutually exclusive events.

$Pr(A\cup B) = Pr(A) + Pr(B) - Pr(A\cap B)$ and $Pr(A\cap B)=0$ whenever $A$ and $B$ are disjoint.

You will multiply probabilities for calculating intersections of events.

$Pr(A\cap B) = Pr(A)\cdot Pr(B|A)$ In the case that $A$ and $B$ are independent events, then $Pr(B|A)=Pr(B)$ in which case the intersection is simply the product $Pr(A)\cdot Pr(B)$.

Related Question