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1) Two bowls each contain 8 pieces of fruit. In bown A there are five oranges and three apples; in bowl B there is one orange and seven apples.
-a- For each bowl, find the probability that two pieces of fruit chosen at random will both be apples.
-b- For each bowl, find the probability that two pieces of fruit chosen at random will both be apples, when the first piece of fruit is replaced before the second is chosen.
-c- One bowl is chosen at random and from it both pieces of fruit are chosen at random without replacement. If both pieces of fruit are apples, find the probability that A was the chosen bowl.
-d- One bowl is chosen at random, and from it two apples are chosen at random, the first apple being replaced before the second is chosen. If both pieces of fruit are apples, find the probability that bowl A was the bowl chosen.
-A-
The probability of both fruits being apples is equal to the probability of drawing an apple the first time and then doing so again the second time.
Bowl A: 3/8 * 2/7 = 3/28
Bowl B: 7/8 * 6/7 = 3/4
-B-
Same procedure as part A, but because the fruit is being replaced, the chances of drawing an apple the second time is the same as drawing an apple the first time.
Bowl A: 3/8 * 3/8 = 9/64
Bowl B: 7/8 * 7/8 = 49/64
-C-
Think about this one logically for a moment. If Bowl A has a 50% chance of drawing two apples and Bowl B has a 25% chance, what would your answer be? The probability of the bowl being Bowl A would be
twice as much, right? That is, a 2/3 chance of the bowl being A, and a 1/3 chance of the bowl being B.
The mathematical form of this logic is:
Chance of Bowl A = Bowl A's chance of drawing two apples / (Bowl A's chance of drawing two apples + Bowl B's chance of drawing two apples).
Chance of Bowl B = 1 - Chance of Bowl A.
Because this is without replacement, we use the figures from part A.
Bowl A: (3/28) / (3/28 + 3/4) = 1/8
Bowl B: 1 - 1/8 = 7/8
-D-
Same idea as part C, except now the fruits are being replaced.
Bowl A: (9/64) / (9/64 + 49/64) = 9/58
Bowl B: 1 - 9/58 = 49/58
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How do you differentiate e to the power of x to the power of (3-x)?
We have not learnt how to deferentiate x with x as the power.
Because e is your base, you don't need to do anything fancy just because you have x as the power. Because it is a composite function, this is simply applying the chain rule.
(e^(3-x))' = e^(3-x) * (3-x)' = -e^(3-x)