Question

A sign on the pumps at a gas station encourages customers to have their oil checked, and claims that one out of 10 cars needs to have oil added. If this is true (and assuming cars arriving independently), find the probability of each of the following:A. One out of the next four cars needs oil.B. Two out of the next eight cars needs oil.C. 5 out of the next 20 cars needs oil.

Answer 1

Answer:

0.2916, 0.1488, 0.0319

Step-by-step explanation:

Given that a sign on the pumps at a gas station encourages customers to have their oil checked, and claims that one out of 10 cars needs to have oil added.

Since each trial is independent there is a constant probability for any random car to need oil is 0.10

Let X be the number of cars that need oil

A) Here X is BIN(4,0.1)

$P(X=1) = 4C1(0.1)(0.9)^3 \\= 0.2916$

B) Here X is Bin (8, 0.1)

$P(x=2) = 8C2 (0.1)^2 (0.9)^6\\\\=0.1488$

C) Here X is Bin (20,5)

$P(x=5) = 20C5 (0.1)^5 (0.9)^{15} \\=0.0319$

Answer 2

This question is about the concept of binomial probability distribution.

A) P(1) = 0.2916

B) P(2) = 0.1488

C) P(5) = 0.03192

• Formula for binomial probability distribution is;

P(x) = nCx • p^(x) • q^(n - x)

Where; q = 1 - p

• From the question, we are told that the claim is that 1 out of 10 cars need to have their oil checked.

Thus; p = 1/10 = 0.1

q = 1 - 0.1

q = 0.9

• A) P(one out of the next four cars need oil) is;

P(1) = 4C1 × 0.1^(1) × 0.9^(4 - 1)

P(1) = 4 × 0.1 × 0.9³

P(1) = 0.2916

• B) P(2 out of the next 8 cars needs oil) is;

P(2) = 8C2 × 0.1² × 0.9^(8 - 2)

P(2) = 28 × 0.01 × 0.531441

P(2) = 0.1488

• C) P(five out of the next twenty cars needs oil) ;

P(5) = 20C5 × 0.1^(5) × 0.9^(20 - 5)

P(5) = 15504 × 0.00001 × 0.20589113209

P(5) = 0.03192

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