Question

In: Statistics and Probability

1. Consider the following events for a college student being selected at random. A = student...

1. Consider the following events for a college student being selected at random. A = student is a hockey player B = student is majoring in kinesiology. Translate the following phrase into symbols: The probability that the student is not a hockey player and is majoring in kinesiology.

Select one:

a. P(AcandB)

b. P(AandBc)

c. P(Bc|A)

d. P(A|Bc)

2. Suppose you roll two fair dice, one that is purple, and one other is violet. Each die can have numbers from 1 to 6. Determine the probability of getting a sum of 5, which means you add the number from each die together obtain the sum of 5.

Select one:

a. 4/36

b. 2/36

c. 5/36

d. 3/36

3. Suppose you have a jar that includes six balls, two gold balls, three purple balls, and one orange ball. You will draw the balls without replacement. Determine the probability of getting a gold ball on the first draw and a purple ball on the second draw. Note that the following probabilities are expressed in terms of products of non-reduced fractions.

Select one:

a. (2/6)(2/6)(2/6)(2/6)

b. (3/6)(3/6)(3/6)(3/6)

c. (2/6)(3/5)(2/6)(3/5)

d. (3/6)(2/5)

Solutions

Expert Solution

(1) P(Not playing Hockey) = 1 - P(playing Hockey) = P(Ac)

P(Student is majoring in kinesiology) = P(B)

The question is asking for the intersection value, and therefore Option a: P(Ac and B)

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(2) Probability = Favorable outcomes / Total Outcomes

We can get a sum of 5 in the following manner, assuming the first is the value on the purple and second is the value on violet. (4,1) (1,4) (2,3) and (3,2) = 4 possible Outcomes

Total outcomes = 62 = 36 (if you throw n dice, the total outcomes = 6n)

Therefore the required probability is Option a: 4 / 36

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(3) Total number of balls = 6

When picking without replacement, from the second pick onwards, the Total outcomes keep reducing by 1, as the first ball picked has not been replaced.

First Draw: P(Gold Ball) = 2 / 6

Second Draw: P(Purple Ball) = 3 / 5

Therefore the required probability is Option d: (3/6) * (2/5)

(It doesn't matter if the figures are changed in the options as the answer still remains the same (2/6) * (3/5) = 1/5 and (3/6) * (2/5) is also = 1/5)

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