Question 1171586: In a bag of gummy worms, 2 are sour and 4 are sweet.
I select a gummy worm and eat it.
I then select a 2nd gummy worm and eat it.
Find the probability that at least 1 of the gummy worms that I eat is sweet.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
There are two possible scenarios here:
Case A) Both gummy worms are sour, or,
Case B) At least one gummy worm is sweet
Only one case can happen. Both cases cannot happen simultaneously. As you can see, case A can be rephrased into "both gummy worms are not sweet".
Let's calculate the probability of case A
We have 2 sour + 4 sweet = 6 gummy worms total.
The probability of getting a sour worm is 2/6 = 1/3.
After selecting that first sour worm, we have 2-1 = 1 sour worm left out of 6-1 = 5 gummies left over. The probability of getting a second sour worm, given the first one was sour, is 1/5.
Multiply the probabilities:
P(both are sour) = P(1st sour)*(2nd sour, given 1st sour)
P(both are sour) = (1/3)*(1/5)
P(both are sour) = 1/15
So we can shorten this to
P(A) = 1/15
to indicate "The probability case A occurs is 1/15".
From here we use the idea that
P(A) + P(B) = 1
because only event A or event B can happen.
So,
P(A) + P(B) = 1
P(B) = 1 - P(A)
P(B) = 1 - 1/15
P(B) = 15/15 - 1/15
P(B) = (15-1)/15
P(B) = 14/15
The probability of getting at least 1 sweet gummy worm is 14/15.
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Here's another approach:
The phrasing "at least one sweet gummy worm" means "one more more sweet gummy worms". So we could select exactly 1 sweet gummy worm, or we could select 2 sweet gummy worms.
First lets calculate the probability of getting exactly 1 sweet gummy worm. We need to find the probability of getting the first one is sweet and the second is sour. Then we need to calculate the first is sour and the second is sweet.
We'll calculate each separately, then add up the results.
P(1st sweet,2nd sour) = P(1st sweet)*P(2nd sour given 1st sweet)
P(1st sweet,2nd sour) = (4/6)*(2/5)
P(1st sweet,2nd sour) = 8/30
P(1st sweet,2nd sour) = 4/15
And,
P(1st sour, 2nd sweet) = (2/6)*(4/5)
P(1st sour, 2nd sweet) = 8/30
P(1st sour, 2nd sweet) = 4/15
We have this nice symmetry going on.
So,
P(exactly 1 sweet) = P(1st sweet,2nd sour) + P(1st sour, 2nd sweet)
P(exactly 1 sweet) = 4/15 + 4/15
P(exactly 1 sweet) = 8/15
Now we have to calculate the probability of getting exactly 2 sweet gummy worms
P(2 sweet) = P(1st sweet)*(2nd sweet given 1st sweet)
P(2 sweet) = (4/6)*(3/5)
P(2 sweet) = 12/30
P(2 sweet) = 6/15
Finally, we add up the results like so
P(at least 1 sweet) = P(exactly 1 sweet)+P(2 sweet)
P(at least 1 sweet) = 8/15 + 6/15
P(at least 1 sweet) = (8+6)/15
P(at least 1 sweet) = 14/15
As you can see, this section was a bit more lengthy. I recommend using the first method.
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Answer: 14/15
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