Binomial Probabilities

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Questions 1-2

179

Question 1.

What is the value of 6!?

Correct. 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720.

Incorrect. 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720.

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Question 2.

How many ways are there to get 6 heads in 10 flips of a coin?

1,000,000

210

Correct. There are

$\begin{bmatrix}10 \\6 \end{bmatrix}$ possible ways. This is

$\frac{10!}{6!(10-6)!}$ = 210.

Incorrect. There are

$\begin{bmatrix}10 \\6 \end{bmatrix}$ possible ways. This is

$\frac{10!}{6!(10-6)!}$ = 210.

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Question 3

416

Question 3.

If n = 8 and p = 0.6, what is P(X = 6)?

8*7*0.6⁶*0.4²

0.6⁶*0.4²

4*7*0.6⁶*0.4²

8*6*0.6⁶*0.4²

Correct. The probability is

$\begin{bmatrix}8 \\6 \end{bmatrix}$ 0.6⁶(1-0.6)²=

$\dfrac{8\,x\, 7\, x\, 6\, x\, 5\, x\, 4\, x\, 3\, x\, 2\, x\, 1}{6\, x\, 5\, x\, 4\, x\, 3\, x\, 2\, x\, 1(2\, x\, 1)}\,0.6(0.4)$ ². Note that 6\,x\,5\,x\,4\,x\,3\,x\,2\,x\,1 cancels from the numerator and denominator and the leftover 2 in the denominator cancels with the 8 to leave 4.

Incorrect. The probability is

$\begin{bmatrix}8 \\6 \end{bmatrix}$ 0.6⁶(1-0.6)²=

$\dfrac{8 \,x \,7 \,x \,6 \,x \,5 \,x \,4 \,x \,3 \,x \,2 \,x \,1}{6 \,x \,5 \,x \,4 \,x \,3 \,x \,2 \,x \,1(2 \,x \,1)}\,0.6(0.4)$ ². Note that 6x5x4x3x2x1 cancels from the numerator and denominator and the leftover 2 in the denominator cancels with the 8 to leave 4.

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Questions 4-7

640

Question 4.

According to government data, about 78% of U.S. households have an automatic dishwasher. If 15 households are selected at random, what is the probability that 9 of them have an automatic dishwasher? Enter your answer using three decimal places.

Try again.

Correct. If X is the number in the sample with an automatic dishwasher, P(X = 9) =

$\begin{bmatrix}15 \\9 \end{bmatrix}$ (0.78)⁹(1-0.78)^15-9.

If X is the number in the sample with an automatic dishwasher, P(X = 9) =

$\begin{bmatrix}15 \\9 \end{bmatrix}$ (0.78)⁹(1-0.78)^15-9.

Question 5.

At a large university, 23% of students come from out-of-state. If we sample 10 students at random, what is the probability that at most 1 is from out-of-state?

0.248

0.219

0.053

0.292

Correct. The chance that at most one of the ten is from out-of-state is P(X

$\leq$ 1) = P(X = 0) + P(X = 1). Using the formula, this becomes

$\begin{bmatrix}15 \\0 \end{bmatrix}$ (0.23)⁰(1-0.23)^10-0 +

$\begin{bmatrix}15 \\1 \end{bmatrix}$ (0.23)¹(1-0.23)^10-1= 0.073 + 0.219.

Incorrect. The chance that at most one of the ten is from out-of-state is P(X

$\leq$ 1) = P(X = 0) + P(X = 1). Using the formula, this becomes

$\begin{bmatrix}15 \\0 \end{bmatrix}$ (0.23)⁰(1-0.23)^10-0 +

$\begin{bmatrix}15 \\1 \end{bmatrix}$ (0.23)¹(1-0.23)^10-1= 0.073 + 0.219.

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Question 6.

Manufacturers don’t like defective products because it can give their brands a bad reputation. However, no production line is perfect. One line is filling soda bottles that are supposed to contain 2 liters. Typically, 99% of these bottles are properly filled. In a quality control sample of 25 bottles, what is the probability that at least 23 are properly filled?

0.024

0.998

0.974

Correct. The probability at least 23 of the 25 are properly filled is P(X

$\geq$ 23), where X is the number of properly filled bottles. Note that if there is a 1% chance of a bottle being underfilled, there is a 99% chance the bottle was properly filled. The desired probability is

$\begin{bmatrix}25 \\23 \end{bmatrix}$ (0.99)²³(1-0.99)^25-23 +

$\begin{bmatrix}25 \\24 \end{bmatrix}$ (0.99)²⁴(1-0.99)^25-24 +

$\begin{bmatrix}25 \\25 \end{bmatrix}$ (0.99)²⁵(1-0.99)^25-25 = 0.024 + 0.196 + 0.778.

Incorrect. The probability at least 23 of the 25 are properly filled is P(X

$\begin{bmatrix}25 \\23 \end{bmatrix}$ (0.99)²³(1-0.99)^25-23 +

$\begin{bmatrix}25 \\24 \end{bmatrix}$ (0.99)²⁴(1-0.99)^25-24+

$\begin{bmatrix}25 \\25 \end{bmatrix}$ (0.99)²⁵(1-0.99)^25-25 = 0.024 + 0.196 + 0.778.

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Question 7.

At a large university, 23% of students come from out-of-state. If we sample 10 students at random, what is the probability at least one of them 1 is from out-of-state?

0.945

0.219

0.292

0.927

Correct. The chance that at least one of the ten is from out-of-state is P(X

$\geq$ 1) = 1 - P(X =0). This becomes 1 -

$\begin{bmatrix}15 \\0 \end{bmatrix}$ (0.23)⁰(1-0.23)^10-0= 1 - 0.073.

Incorrect. The chance that at least one of the ten is from out-of-state is P(X

$\geq$ 1) = 1 - P(X =0). This becomes 1 -

$\begin{bmatrix}15 \\0 \end{bmatrix}$ (0.23)⁰(1-0.23)^10-0= 1 - 0.073.

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StatTutor Lesson - Binomial Probabilities

Questions 1-2

Question 1.

Question 2.

Question 3

Question 3.

Questions 4-7

Question 4.

Question 5.

Question 6.

Question 7.