Let T be the event that an adult admits to texting while driving and N be the event an adult does not admit to texting while dri
ving. We previously determined P(T) = 0.61
and
P(N) = 0.39.
Since three adults are chosen randomly, we have the following simple events.
TTT TTN TNT NTT TNN NTN NNT NNN
The adults were randomly selected, indicating these can be seen as independent events. Therefore, the multiplication rule can be used. Recall the multiplication rule states that for independent events, the probability that they all occur is the product of their respective probabilities. Let x be the number of adults who admit to texting while driving. Since three adults are randomly selected, then x can take on the values 0, 1, 2, or 3.
When x = 0, then no adult in the group of three admits to texting while driving. This corresponds to the simple event NNN whose probability is calculated as below.
P(x = 0) = P(NNN)
= P(N)P(N)P(N)
= 0.39(0.39)(0.39)
=
When x = 1, then only one adult in the group admits to texting while driving. This corresponds to the simple events TNN, NTN, and NNT. First, calculate the probability of each simple event by multiplying the individual probabilities. Then sum the three simple events to find
P(x = 1).
Calculate
P(x = 1).
P(x = 1) = P(TNN) + P(NTN) + P(NNT)
= P(T)P(N)P(N) + P(N)P(T)P(N) + P(N)P(N)P(T)
= 0.61(0.39)(0.39) + 0.39(0.61)(0.39) + 0.39(0.39)(0.61)
=
Find the remaining probabilities
P(x = 2)
and
P(x = 3).
P(x = 2) = P(TTN) + P(TNT) + P(NTT)
= P(T)P(T)P(N) + P(T)P(N)P(T) + P(N)P(T)P(T)
= 0.61(0.61)(0.39) + 0.61(0.39)(0.61) + 0.39(0.61)(0.61)
=
P(x = 3) = P(TTT)
= P(T)P(T)P(T)
= 0.61(0.61)(0.61)
=
