Chapter 5, Problem R5.6RE

(a)

To determine

Whether the events “thick − crust pizza” and “pizza without mushroom” are mutually exclusive.

Answer to Problem R5.6RE

The events “thick − crust pizza” and “pizza without mushroom” are not mutually exclusive.

Explanation of Solution

Given information:

There are total 9 pizzas.

Such that

3 of the 9 pizzas have thick crust,

2 of the 3 thick − crust pizzas have mushrooms,

And

4 of the 6 thin − crust pizzas have mushrooms.

The two events are mutually exclusive (or disjoint), if the events cannot occur at same time.

According to the statement,

There are 2 thick − crust pizzas with mushroom.

For the events to be truly mutually exclusive,

There needs to be 0 thick − crust pizzas.

Thus,

The events “thick − crust pizza” and “pizza without mushroom” are not mutually exclusive.

(b)

To determine

Whether the events “thick − crust pizza” and “pizza without mushroom” are independent.

Answer to Problem R5.6RE

The events “thick − crust pizza” and “pizza without mushroom” are independent.

Explanation of Solution

Given information:

There are total 9 pizzas.

Such that

3 of the 9 pizzas have thick crust,

2 of the 3 thick − crust pizzas have mushrooms,

And

4 of the 6 thin − crust pizzas have mushrooms.

Two events are independent, if the probability of occurrence of one event does not affect the probability of occurrence of other event.

Note that

3 of the 9 pizzas have thick − crust.

In this case, the number of favorable outcomes is 3 and number the number possible outcomes is 9.

If we divide number of favorable outcomes by the number of possible outcomes, we get the probability.

Thus,

The probability for first mushroom pizza,

  $P\left(\text{Thick}-\text{crust}\text{pizza}\right)=\frac{\text{Number}\text{of}\text{favorable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{3}{9}=\frac{1}{3}$

Now,

According to statement,

6 pizzas have mushroom.

Whereas,

2 of these 6 pizzas are thick − crust pizzas.

Thus,

The conditional probability for pizza with mushrooms have thick − crust,

  $P\left(\text{Thick}-\text{crust}\text{pizza}|\text{pizza}\text{with}\text{mushrooms}\right)=\frac{\text{Number}\text{of}\text{favorable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{2}{6}=\frac{1}{3}$

From above two probabilities,

We came to know that

Both probabilities are equal.

This implies

Pizza having mushrooms or not does not affect the probability of having a thick − crust pizza.

Thus,

The events are independent.

(c)

To determine

Probability for the randomly selected 2 pizzas has mushrooms.

Answer to Problem R5.6RE

Probability that both pizzas have mushroom is approx. 0.4167.

Explanation of Solution

Given information:

There are total 9 pizzas.

Such that

3 of the 9 pizzas have thick crust,

2 of the 3 thick − crust pizzas have mushrooms,

And

4 of the 6 thin − crust pizzas have mushrooms.

General multiplication rule:

  $P\left(A\text{and}B\right)=P\left(A\cap B\right)=P\left(A\right)\times P\left(B|A\right)=P\left(B\right)\times P\left(A|B\right)$

Note that

6 of the 9 pizzas have mushrooms.

In this case, the number of favorable outcomes is 6 and number the number possible outcomes is 9.

If we divide number of favorable outcomes by the number of possible outcomes, we get the probability.

Thus,

The probability for first mushroom pizza,

  $P\left(\text{First}\text{mushroom}\text{pizza}\right)=\frac{\text{Number}\text{of}\text{favorable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{6}{9}$

Since one mushroom pizza has already been selected, we are left with 5 pizzas have mushrooms out of remaining 8 pizzas.

In this case, the number of favorable outcomes is 5 and number the number possible outcomes is 8.

If we divide number of favorable outcomes by the number of possible outcomes, we get the probability.

Thus,

Without replacement, the probability for second mushroom pizza after first mushroom pizza,

  $P\left(\text{Second}\text{mushroom}\text{pizza}|\text{First}\text{mushroom}\text{pizza}\right)=\frac{\text{Number}\text{of}\text{favorable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{5}{8}$

Apply general multiplication rule:

  $\begin{array}{l}P\left(\text{Two}\text{mushrooms}\text{pizzas}\right)=P\left(\text{First}\text{mushroom}\text{pizza}\right)\times P\left(\text{Second}\text{mushroom}\text{pizza}|\text{First}\text{mushroom}\text{pizza}\right)\\ =\frac{6}{9}\times \frac{5}{8}\\ =\frac{30}{72}\\ =\frac{5}{12}\\ \approx 0.4167\end{array}$

Thus,

Probability for randomly selected 2 pizzas has mushroom is approx. 0.4167.