Question

A box contains 4 red and 5 blue marbles. Michele picks three marbles at random from this box. If Z is a random variable representing the # of blue marbles picked from the box, do the following:
a. Express the probability mass function of Z in tabular form.
b. Draw the corresponding histogram.
c. Compute the probability that Michele can pick more red marbles than blue from the box.

Answer 1

Answer on Question #72678, Math / Statistics and Probability

A box contains 4 red and 5 blue marbles. Michele picks three marbles at random from this box. If $Z$ is a random variable representing the # of blue marbles picked from the box, do the following:

a. Express the probability mass function of $Z$ in tabular form.

b. Draw the corresponding histogram.

c. Compute the probability that Michele can pick more red marbles than blue from the box.

Solution

a. Express the probability mass function of $Z$ in tabular form.

Let $Z =$ number of times the blue marble is picked from the box. The 8 possible elementary events and the corresponding values for $Z$ , are

If 3 three marbles are picked at random from the box, what is the probability that none is blue?

$$P (R R R) = \frac {4}{9} \cdot \frac {3}{8} \cdot \frac {2}{7} = \frac {1}{2 1}$$

If 3 three marbles are picked at random from the box, what is the probability that one marble is blue?

$$P (R R B) + P (R B R) + P (B R R) = \frac {4}{9} \cdot \frac {3}{8} \cdot \frac {5}{7} + \frac {4}{9} \cdot \frac {5}{8} \cdot \frac {3}{7} + \frac {5}{9} \cdot \frac {4}{8} \cdot \frac {3}{7} = \frac {5}{1 4}$$

If 3 three marbles are picked at random from the box, what is the probability that two marbles are blue?

$$P (R B B) + P (B B R) + P (B R B) = \frac {4}{9} \cdot \frac {5}{8} \cdot \frac {4}{7} + \frac {5}{9} \cdot \frac {4}{8} \cdot \frac {4}{7} + \frac {5}{9} \cdot \frac {4}{8} \cdot \frac {4}{7} = \frac {1 0}{2 1}$$

If 3 three marbles are picked at random from the box, what is the probability that three marbles are blue?

$$P (B B B) = \frac {5}{9} \cdot \frac {4}{8} \cdot \frac {3}{7} = \frac {5}{4 2}$$

Check

$$\frac {1}{2 1} + \frac {5}{1 4} + \frac {1 0}{2 1} + \frac {5}{4 2} = \frac {2 + 1 5 + 2 0 + 5}{4 2} = 1$$

Therefore, the probability distribution for the number of heads occurring in three coin tosses is:

b. Draw the corresponding histogram.

Probability Mass Function

c. Compute the probability that Michele can pick more red marbles than blue from the box.

This is the probability of picking either 2 or 3 red marbles (0 or 1 blue marbles)

$$P(RRB) + P(RBR) + P(BRR) + P(RRR) = \frac{5}{14} + \frac{1}{21} = \frac{17}{42}$$

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