Question

There are two types of cellular phones, handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles. Phone calls can be classified by the traveling speed of the user as fast (F) or slow (W). Monitor a cellular phone call and observe the type of telephone and the speed of the user. The probability model for this experiment has the following information: P[F]=0.5, P[HF]=0.2, P[MW]=0.1. What is the sample space of the experiment? Find the following probabilities P[W], P[MF], and P[H].

Solution

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The sample space contains 4 two-letter words with the first letter either H (Handheld phones) or M (Mobile phones mounted in vehicles) and the second letter either F (Fast speed of the user) or W (Slow speed of the user). For example, if an element of the sample space is MF, it refers to a phone call on a mobile phone Mounted in a vehicle which is running Fast.

Hence, Sample Space, S = {HF, HW, MF, MW}\boxed{\textbf{S = \{HF, HW, MF, MW\}}}

Given: P(F) = 0.5, P(HF) = 0.2, P(MW) = 0.1

Since we can see that a phone call can either be fast or slow, {F, W} is a partition.

Hence, P(F) + P(W) = 1.0 \ \ \ (by properties of partition)

Hence, P(W) + 0.5 = 1.0

Hence, P(W) = 1.0 - 0.5 = 0.5\boxed{\textbf{0.5}}

We a phone call can be either on a handheld phone or a mobile phone mounted in vehicle, {H, M} is a partition.

Hence, P(H \cap F) + P(M \cap F) = P(F)

Hence, P(F) = P(HF) + P(MF)

Hence, 0.5 = 0.2 + P(MF)

Hence, P(MF) = 0.5 - 0.2 = 0.3\boxed{\textbf{0.3}}

Again, since a phone call can either be Fast or Slow, {F, W} is a partition.

Hence, P(M \cap F) + P(M \cap W) = P(M)

Hence, P(M) = P(MF) + P(MW)

Hence, P(M) = 0.3 + 0.1 = 0.4

Now, since phone call can be either on a handheld phone or a mobile phone mounted in a vehicle, {H, M} is a partition.

Hence, P(H) + P(M) = 1.0

Hence, P(H) + 0.4 = 1.0

Hence, P(H) = 1.0 - 0.4 = 0.6\boxed{\textbf{0.6}}

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