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Andreas93 [3]
3 years ago
9

Amy, Bess, Cassie, Diana, and Ellen are a basketball team. Two of them will be guards. Assuming they all have an equal chance of

becoming guards, what is the probability that Diana will be a guards. which outcome (or outcomes) of the sample space compose the event? Describe the probability Diana becoming a guard as impossiable, unlikey, neither likey nor unlikey, likey or certain. Justify your answ
Mathematics
2 answers:
Vinvika [58]3 years ago
5 0
There are 5 people who can be guards, 2 will be guards. Since each person has the same chance of becoming a guard, each person has a 2/5 chance of being a guard which makes it unlikely for Diana to become a guard.
alexgriva [62]3 years ago
5 0

Answer: 1: The sample space for this problem is {Amy, Bess, Cassie, Diana, Ellen} 2: 10 outcomes 3: 2/5 or 0.4 4: The probability of Diana becoming a guard is possible since her chance of being chosen is 40%.

Step-by-step explanation: The explanations are on every problem except for 2 so here it is. The outcomes of the sample problem are Diana and Ellen, Diana and Cassie, Diana and Amy, Diana and Bess, Amy and Bess, Amy and Cassie, Amy and Ellen, Cassie and Bess, Cassie and Ellen, Ellen and Bess which adds up to ten outcomes since two people can be chosen from the group of five.

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4 0
3 years ago
Problem of the Day:
Shtirlitz [24]

Answer:

Step-by-step explanation:

What this question is asking of you is what is the greatest common divisor of 12 and 15. Or, what is the biggest number that divides both 12 and 15.

in order to find this we have to split each number into it's prime components.

for 12 they are 2,2 and 3 (

2

⋅

2

⋅

3

=

12

)

and for 15 they are 3 and 5 (

3

⋅

5

=

15

)

Out of those two groups (2,2,3) and (3,5) the only thing in common is 3, so 3 is the greatest common divisor. That tells us that the greatest number of groups that can exist and have the same number of girls and the same number of boys for each group is 3.

Now to find out how many girls and boys there are going to be in each group we divide the totals by 3, so:

12

3

=

4

girls per group, and

15

3

=

5

boys per group.

(just as a thought exercise, if there were 16 boys, the divisors would have been (2,2,3) and (2,2,2,2), leaving us with 4 groups [

2

⋅

2

] of 3 girls [12/4] and 4 boys [16/4] )

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