Example




An urn contains 4 white marbles and 6 blue marbles.

  1. If two marbles are drawn from the urn, what is the probability that both are blue?
  2. If two marbles are drawn, what is the probability that at least one is blue?
  3. Two marbles are drawn, and the second one is blue. What is the probability that the first one was white?
  4. One marble is drawn from the urn. If it is blue, then one die is tossed. If the marble is white, then a pair of dice are tossed. What is the probability that the total of the (die or) dice is less than 6?


Solution

An urn contains 4 white marbles and 6 blue marbles so there is a total of 10 marbles in the urn.


Part 1)

There are (10C2=) 45 ways to select any two marbles from the urn while there are (6C2=) 15 ways to select two blue marbles from the urn. Therefore, the probability of choosing two blue marble is given as follows. 


If two marbles are drawn from the urn, then the probability that both are blue is 0.3333.


Part 2)

There are (10C2=) 45 ways to select any two marbles from the urn while there are (4C2=) 6 ways to select two white marbles (no blue marbles) from the urn. Therefore, the probability of choosing two white marbles is given as follows.


If two marbles are drawn from the urn, then the probability that both are white is 0.1333. This gives, the probability that at least one is blue = 1 – P(no blue marbles) = 1 – 0.1333 = 0.8667.


Part 3) 

Suppose two marbles are drawn. Initially, there were 10 marbles in the urn. At the first draw, we can either choose a white or a blue marble. So, the probability of getting the white marble at the first draw is simply,


Similarly, the probability of getting a blue marble at the first draw is 6/10 = 0.60.

Now at the second draw, again we can either choose a white or a blue marble from the remaining 9 marbles. Unlike the first draw, the probability of getting a white marble or a blue marble at the second draw is dependent on whether we have drawn a white or a blue marble at the first draw.

This gives, the probability of getting any of the white or a blue marble at the first draw is independent of getting a white or a blue marble at the second draw.

 Therefore, when two marbles are drawn, and the second one is blue. Then the probability that the first one was white is 0.40.


Part 4)

One marble is drawn from the urn. If it is blue, then one die is tossed. If the marble is white, then a pair of dice are tossed. Let P(R) be the probability of getting a white marble and P(B) the probability of getting a blue marble.

From part 3, we get P(R) = 0.40 and P(B) = 0.60.

Let T be the event that the total of the (die or) dice is less than 6.

When a pair of dice is tossed then the probability of getting the total is less than 6 is, 0.2778 (=10/36). While one die is tossed then the probability of getting the total is less than 6 is, 0.8333 (=5/6).

In the context of this problem, the probability of getting total is less than 6 when the white marble is drawn is, P(T | R) = 0.2778, and the probability of getting total is less than 6 when the blue marble is drawn is, P(T | B) = 0.8333.


Required probability

The probability that the total of the (die or) dice is less than 6, that is, P(T).


Law of total probability 

P(T) = P(T | R)*P(R) + P(T | B)*P(B)


Using the law of total probability, P(T) = 0.2778*0.40 + 0.8333*0.60 = 0.6111.

Therefore, the probability that the total of the (die or) dice is less than 6 is 0.6111.