Question

integrated ii: conditional probability
conditional probability
mit opencourseware
open - ended question
what is the probability of getting a blue marble from both bowls?

Explanation:

Step1: Assume probabilities for each bowl

Let the probability of getting a blue marble from the first bowl be $P_1$ and from the second bowl be $P_2$. Since no information about the contents of the bowls is given, if we assume the first bowl has $n_1$ marbles and $b_1$ blue - marbles, then $P_1=\frac{b_1}{n_1}$, and for the second bowl with $n_2$ marbles and $b_2$ blue - marbles, $P_2 = \frac{b_2}{n_2}$.

Step2: Use multiplication rule for independent events

If the events of drawing a blue marble from the first bowl and from the second bowl are independent, the probability of getting a blue marble from both bowls is $P = P_1\times P_2=\frac{b_1}{n_1}\times\frac{b_2}{n_2}$. Without specific values for $n_1,b_1,n_2,b_2$, we can't give a numerical answer.

Answer:

We need more information about the number of blue marbles and total marbles in each bowl to calculate the probability.