Chapter 5.3, Problem 92E

To determine

Probability for getting at least one 6.

Answer to Problem 92E

Probability for getting at least one 6 is 0.8385.

Explanation of Solution

Given information:

Fair, six − sided die rolled 10 times.

Such that

At least 1 of the 10 rolls results in a 6.

Calculations:

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

According to multiplication rule for independent events,

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

According to complement rule,

  $P\left({\text{A}}^{\text{c}}\right)=P\left(\text{not}\text{A}\right)=1-P\left(\text{A}\right)$

Let

A: One roll results in a 6

B: At least 1 of the 10 rolls results in a 6

A^c: One roll does not result in a 6

B^c: None of the 10 rolls results in a 6

We know that

Fair − sided die has 6 possible outcomes. i.e. 1, 2, 3, 4, 5, 6.

Since each outcome is equally possible, we have one chance in six possible outcomes to obtain a 6 on one roll.

Thus, the number of favorable outcome is 1, and number of possible outcomes is 6.

When the number of favorable outcomes is divided by the number of possible outcomes, we get the probability.

  $P\left(\text{A}\right)=\frac{\text{Number}\text{of}\text{favorable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{1}{6}$

Now,

Apply the complement rule:

  $P\left({\text{A}}^{\text{c}}\right)=1-P\left(\text{A}\right)=1-\frac{1}{6}=\frac{5}{6}$

Since die has been rolled 10 times, it would be more convenient to assume that each roll is independent.

Thus,

For probability that none of the 10 rolls results in a 6 apply multiplication rule for independent events:

  $P\left({\text{B}}^{\text{c}}\right)=\underset{10\text{repetitions}}{\underbrace{P\left({\text{A}}^{\text{c}}\right)\times P\left({\text{A}}^{\text{c}}\right)\times \mathrm{...}\times P\left({\text{A}}^{\text{c}}\right)}}={\left(P\left({\text{A}}^{\text{c}}\right)\right)}^{10}={\left(\frac{5}{6}\right)}^{10}=\frac{9,765,625}{60,466,176}\approx 0.1615$

Apply the complement rule:

  $P\left(\text{B}\right)=P\left({\left({\text{B}}^{\text{c}}\right)}^{\text{c}}\right)=1-P\left({\text{B}}^{\text{c}}\right)=1-0.1615=0.8385$

Thus,

Probability that at least 1 of the 10 rolls results in a 6 is 0.8385.