2. A candy machine has 10 grape, 6 green apple, and 16 mango gumballs. What is the probability of selecting a grape and a mango gumball as fraction in simplest form and a percent?
P(A and B) = P (A) * P (B)
10/32 * 16/32 = 80/512 = 5/32 simplest form
Percent: 15.625%
The probability of choosing a grape and a mango gumball is 5/32 or 15.625%.
P(A and B) = P (A) * P (B)
10/32 * 16/32 = 80/512 = 5/32 simplest form
Percent: 15.625%
The probability of choosing a grape and a mango gumball is 5/32 or 15.625%.
Answers
Answered by
Damon
That depends :(
If you do not put the grape back, you only have 31 in the machine when you draw the mango.
You assumed the two events were independent. They may not be.
I would have said:
10/32 * 16/31 not 32
If you do not put the grape back, you only have 31 in the machine when you draw the mango.
You assumed the two events were independent. They may not be.
I would have said:
10/32 * 16/31 not 32
Answered by
Anonymous
So it's dependent event because you don't replace it, so instead of having a total of 32 it's 31?
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