# Chapter 1. Binomial Probabilities

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### Question 1.1

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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 1.2

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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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6:56

### Question 1.3

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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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10:40

### Question 1.4

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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 1.5

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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 1.6

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 $$\geq$$ 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.
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.