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%.
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
So it's dependent event because you don't replace it, so instead of having a total of 32 it's 31?
To find the probability of selecting a grape and a mango gumball from the candy machine, we use the formula for the intersection of two independent events:
P(A and B) = P(A) * P(B)
Here, A represents the event of selecting a grape gumball, and B represents the event of selecting a mango gumball.
To find the individual probabilities of each event, we divide the number of gumballs of that type by the total number of gumballs in the machine.
There are 10 grape gumballs out of a total of 32 gumballs, so P(A) = 10/32.
Similarly, there are 16 mango gumballs out of a total of 32 gumballs, so P(B) = 16/32.
To find the probability of both events occurring, we multiply the probabilities of the individual events:
P(A and B) = (10/32) * (16/32) = 160/1024 = 5/32 in simplest form.
The probability of selecting a grape and a mango gumball is therefore 5/32.
To express this probability as a percentage, we divide the numerator (5) by the denominator (32), and multiply by 100:
(5/32) * 100 = 15.625%
So, the probability of selecting a grape and a mango gumball is 5/32 or 15.625%.