Chapter 1. Binomial Probabilities

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.

2

Try again.

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.

2

Try again.

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.

2

Try again.

2

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.

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.

2

Try again.

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.

2

Try again.

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.

2

Try again.