Answer: - 0.28
Explanation:
1) Expected value: is the weighted average of the values, being the probabilities the weight.
That is: ∑ of prbability of event i × value of event i.
In this case: (probability of getting 2 or 12) × (+6) + (probability of gettin 3 or 11) × (+2) + (probability of any other sum) × (-1).
2) Sample space:
Sum Points awarded
1+ 1 = 2 +6
1 + 2 = 3 +2
1 + 3 = 4 -1
1 + 4 = 5 -1
1 + 5 = 6 -1
1 + 6 = 7 -1
2 + 1 = 3 +2
2 + 2 = 4 -1
2 + 3 = 5 -1
2 + 4 = 6 -1
2 + 5 = 7 -1
2 + 6 = 8 -1
3 + 1 = 4 -1
3 + 2 = 5 -1
3 + 3 = 6 -1
3 + 4 = 7 -1
3 + 5 = 8 -1
3 + 6 = 9 -1
4 + 1 = 5 -1
4 + 2 = 6 -1
4 + 3 = 7 -1
4 + 4 = 8 -1
4 + 5 = 9 -1
4 + 6 = 10 -1
5 + 1 = 6 -1
5 + 2 = 7 -1
5 + 3 = 8 -1
5 + 4 = 9 -1
5 + 5 = 10 -1
5 + 6 = 11 +2
6 + 1 = 7 -1
6 + 2 = 8 -1
6 + 3 = 9 -1
6 + 4 = 10 -1
6 + 5 = 11 +2
6 + 6 = 12 +6
2) Probabilities
From that, there is:
- 2/36 probabilities to earn + 6 points.
- 4/36 probabilites to earn + 2 points
- the rest, 30/36 probabilities to earn - 1 points
3) Expected value = (2/36)(+6) + (4/36) (+2) + (30/36) (-1) = - 0.28
Answer: 120
Step-by-step explanation:
From the question, we are informed that there are 10 people in a math club and that three people will be chosen for the Pi Day committee.
The number of different ways that the club members can structure the committee goes thus:
This can be solved by:
nCr = n! / r! (n - r)!
where n = 10
r = 3
= n! / r! (n - r)!
= 10! / 3! (10 - 3)!
= (10 × 9 × 8) / (3 × 2)
=120
The 3rd answer, is correct.
