Mathematics, 25.08.2019 19:50 902coo

If someone could me with this problem it'd be very much *sorry for the quality btw

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Okay, lets write down what we do know...

4 out of the 12 cups are decaf, so the odds of selecting a decaf cup by random is 4/12 or 1/3

Alternatively 8 of the 12 cups are caffeinated, so the odds of selecting a caffeinated cup by random is 8/12 or 2/3

P(decaf)=1/3
P(caf)=2/3

But those odds are only for the first cup picked. The second cup has 4 probabilities...

1. Picking a decaf cup for the second time

2. Picking a decaf cup for the first time

3. Picking a caf cup for the second time

4. Picking a caf cup for the first time

So lets find the probability of each event...

1. Picking a decaf cup for the second time. If a decaf cup has already been picked, then the odds of drawing another one is 3/11

2. Picking a decaf cup for the first time. If a decaf cup hasn't already been picked, then the odds of drawing one is 4/11

3. Picking a caf cup for the second time. If a caf cup has already been picked, then the odds of drawing one is 7/11

4. Picking a caf cup for the first time. If a car cup hasn't already been picked, then the odds of drawing one is 8/11.

Okay, so now we can solve for the various possibilities.

1. Picking a decaf cup for the second time
1/3*3/11=3/33

2. Picking a decaf cup for the first time
2/3*4/11=8/33

3. Picking a caf cup for the second time
2/3*7/11=14/33

4. Picking a caf cup for the first time
2/3*8/11=16/33

Both cups were caffeinated, so neither were decaf.

Answer= Neither cup is decaf.

Answer from: Quest

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Answer from: Quest

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If someone could me with this problem it'd be very much *sorry for the quality btw
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