Question

The most common bet in craps is the “pass line.” A pass line bettor wins immediately if either a $7$or an$11$comes up on the first roll. This is called a natural. What is the probability that a natural does not occur?

a. $2/36$

b.$6/36$

c.$8/36$

d. $16/36$

e. $28/36$

Short answer

The probability that a natural does not occur is (e) $28/36$

Step-by-step solution

Given information

We need to find the probability that a natural does not occur .

Explanation

The probability of rolling a $7$ is $6/36$ , while the probability of rolling an$11$ is $2/36.$

So , we get$P\left(7\right)$= $6/36$, $P\left(11\right)=2/36$

As, it is not possible to roll two different sums, the two events must be mutually exclusive.

We will use the addition rule for mutually exclusive events: P(natural) = $P(7or11)=P\left(7\right)+P\left(11\right)=8/36$

So, P(not natural)=$1-\frac{8}{36}=\frac{28}{36}$