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iris [78.8K]
3 years ago
14

(7 points) Cars arrive at a toll both according to a Poisson process with mean 80 cars per hour. If the attendant makes a two-mi

nute phone call, what is the probability that at least 1 car arrives during the call
Mathematics
1 answer:
saveliy_v [14]3 years ago
5 0

Answer:

The probability that at least 1 car arrives during the call is 0.9306

Step-by-step explanation:

Cars arriving according to Poisson process - 80 Cars per hour

If the attendant makes a 2 minute phone call, then effective λ = 80/60 * 2 = 2.66666667 = 2.67   X ≅ Poisson (λ = 2.67)

Now, we find the probability: P(X≥1)

P(X≥1) = 1 - p(x < 1)

P(X≥1) = 1 - p(x=0)

P(X≥1) = 1 - [ (e^-λ) * λ^0] / 0!

P(X≥1) = 1 - e^-2.67

P(X≥1) = 1 - 0.06945

P(X≥1) = 0.93055

P(X≥1) = 0.9306

Thus, the probability that at least 1 car arrives during the call is 0.9306.

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Among persons donating blood to a clinic, 85% have Rh+ blood (that is, the Rhesus factor is present in their blood.) Six people
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a) There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) There is a 22.36% probability that at most four of the six have Rh+ blood.

c) There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

Step-by-step explanation:

For each person donating blood, there are only two possible outcomes. Either they have Rh+ blood, or they do not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

p = 0.85, n = 6.

a) fine the probability that at least one of the five does not have the Rh factor.

Either all six have the factor, or at least one of them do not. The sum of the probabilities of these events is decimal 1. So:

P(X < 6) + P(X = 6) = 1

P(X < 6) = 1 - P(X = 6)

In which:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771

So

P(X < 6) = 1 - P(X = 6) = 1 - 0.3771 = 0.6229

There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) find the probability that at most four of the six have Rh+ blood.

Either more than four have Rh+ blood, or at most four have. So

P(X \leq 4) + P(X > 4) = 1

P(X \leq 4) = 1 - P(X > 4)

In which

P(X > 4) = P(X = 5) + P(X = 6)

P(X = 5) = C_{6,5}.(0.85)^{5}.(0.15)^{1} = 0.3993

P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771

P(X > 4) = P(X = 5) + P(X = 6) = 0.3993 + 0.3771 = 0.7764

P(X \leq 4) = 1 - P(X > 4) = 1 - 0.7764 = 0.2236

There is a 22.36% probability that at most four of the six have Rh+ blood.

c) The clinic needs six Rh+ donors on a certain day. How many people must donate blood to have the probability of obtaining blood from at least six Rh+ donors over 0.95?

With 6 donors:

P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771

37.71% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 7 donors:

P(X = 6) = C_{7,6}.(0.85)^{6}.(0.15)^{1} = 0.3960

0.3771 + 0.3960 = 0.7764 = 77.64% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 8 donors

P(X = 6) = C_{8,6}.(0.85)^{6}.(0.15)^{2} = 0.2376

0.3771 + 0.3960 + 0.2376 = 1.01 = 101% probability of obtaining blood from at least six Rh+ donors over 0.95.

There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

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3 years ago
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