<span>2/15 if drawn without replacement.
1/9 if drawn with replacement.
Assuming that the chips are drawn without replacement, there are 6 * 5 different possibilities. And that's a low enough number to exhaustively enumerate them. So they are:
1,2 : 1,3 : 1,4 : 1,5 : 1,6
2,1 : 2,3 : 2,4 : 2,5 : 2,6
3,1 : 3,2 : 3.4 : 3,5 : 3,6
4,1 : 4,2 : 4.3 : 4,5 : 4,6
5,1 : 5,2 : 5.3 : 5,4 : 5,6
6,1 : 6,2 : 6.3 : 6,4 : 6,5
Of the above 30 possible draws, there are 4 that add up to 5. So the probability is 4/30 = 2/15
If the draw is done with replacement, then there are 36 possible draws. Once again, small enough to exhaustively list, they are:
1,1 : 1,2 : 1,3 : 1,4 : 1,5 : 1,6
2,1 : 2,2 : 2,3 : 2,4 : 2,5 : 2,6
3,1 : 3,2 : 3,3 : 3.4 : 3,5 : 3,6
4,1 : 4,2 : 4.3 : 4,4 : 4,5 : 4,6
5,1 : 5,2 : 5.3 : 5,4 : 5,5 : 5,6
6,1 : 6,2 : 6.3 : 6,4 : 6,5 : 6,6
And of the above 36 possibilities, exactly 4 add up to 5. So you have 4/36 = 1/9</span>
Answer:
about 11.03 million
Step-by-step explanation:
Use the equation I = P(1+r/100)^n - P (I is the compound interest, P is the principle, r is the rate percent, and n is the number of years):
Substitute the values given:
I = 70,000,000(1 + 5/100)^3 - 70,000,000
Use a calculator to solve and you will get ~11.03 million.
Answer:
P(A) = 0.2
P(B) = 0.25
P(A&B) = 0.05
P(A|B) = 0.2
P(A|B) = P(A) = 0.2
Step-by-step explanation:
P(A) is the probability that the selected student plays soccer.
Then:
P(B) is the probability that the selected student plays basketball.
Then:
P(A and B) is the probability that the selected student plays soccer and basketball:
P(A|B) is the probability that the student plays soccer given that he plays basketball. In this case, as it is given that he plays basketball only 10 out of 50 plays soccer:
P(A | B) is equal to P(A), because the proportion of students that play soccer is equal between the total group of students and within the group that plays basketball. We could assume that the probability of a student playing soccer is independent of the event that he plays basketball.
