A box contains 4 pears and 7 Oranges. Three fruited are taken out at random and eaten. Find the probability that (1) 2 pears and 1 orange are eaten (2) The third fruit is eaten is an orange (3) The first fruit eaten was a pears. Given that the third fruit eaten is an orange

(1) Probability of 2 pears and 1 orange being eaten:

To calculate this, we need to find the probability of choosing 2 pears out of 4 pears, multiplied by the probability of choosing 1 orange out of 7 oranges. Since the order doesn't matter, we also need to consider the possible ways these fruits can be chosen.

Probability of choosing 2 pears out of 4:
C(4, 2) represents the number of ways to choose 2 pears out of 4, which is the combination of 4 objects taken 2 at a time.

C(4, 2) = 4! / (2! * (4 - 2)!) = 6

Probability of choosing 1 orange out of 7:
C(7, 1) = 7! / (1! * (7 - 1)!) = 7

Number of ways to arrange 2 pears and 1 orange:
Since the order doesn't matter, we can choose any 3 fruits out of the 11 total fruits.

C(11, 3) = 11! / (3! * (11 - 3)!) = 165

Now we can calculate the probability:

P(2 pears and 1 orange) = (C(4, 2) * C(7, 1)) / C(11, 3)
= (6 * 7) / 165
= 42/165
= 14/55

So, the probability of eating 2 pears and 1 orange is 14/55.

(2) Probability that the third fruit eaten is an orange:
Since the first two fruits are already eaten, we only need to consider the remaining fruits in the box. There are 8 fruits remaining, with 4 being oranges.

Probability of choosing an orange:
C(4, 1) = 4! / (1! * (4 - 1)!) = 4

Probability of choosing any fruit out of the remaining 8:
C(8, 1) = 8! / (1! * (8 - 1)!) = 8

Now we can calculate the probability:

P(third fruit is an orange) = C(4, 1) / C(8, 1)
= 4/8
= 1/2

So, the probability of the third fruit being an orange is 1/2.

(3) Probability that the first fruit eaten was a pear, given that the third fruit eaten is an orange:

To calculate this, we need to consider the remaining fruits after the third fruit is eaten. Since there are 8 remaining fruits, with 4 being oranges, and we know that the third fruit is an orange, we need to find the probability that the first fruit eaten, out of the remaining 10 fruits, is a pear.

Probability of choosing a pear:
C(4, 1) = 4! / (1! * (4 - 1)!) = 4

Probability of choosing any fruit out of the remaining 10:
C(10, 1) = 10! / (1! * (10 - 1)!) = 10

Now we can calculate the probability:

P(first fruit eaten was a pear, given third fruit is an orange) = C(4, 1) / C(10, 1)
= 4/10
= 2/5

So, the probability that the first fruit eaten was a pear, given that the third fruit eaten is an orange, is 2/5.

To find the probability in each scenario, we first need to determine the total number of possible outcomes.

1) Total number of possible outcomes:
Since there are 11 fruits in total, the number of possible outcomes is given by the combination formula: C(11,3) = 165.

(1) Probability of eating 2 pears and 1 orange:
To calculate the probability of this event, we need to determine the number of favorable outcomes. There are 4 pears to choose from, and we need to select 2 pears out of the 4: C(4,2) = 6.
Similarly, we have 7 oranges to choose from and we need to choose 1 orange out of the 7: C(7,1) = 7.
Therefore, the number of favorable outcomes is 6 * 7 = 42.

Probability = Number of favorable outcomes / Total number of possible outcomes = 42 / 165 ≈ 0.2545

(2) Probability of the third fruit eaten being an orange:
Since there are 11 fruits in total, and we are choosing 3 fruits at random without replacement, there are 11 choices for the first fruit, 10 choices for the second fruit, and 9 choices for the third fruit.
The number of favorable outcomes in this case is 7 (since we want the third fruit eaten to be an orange).

Probability = Number of favorable outcomes / Total number of possible outcomes = 7 / 165 ≈ 0.0424

(3) Probability that the first fruit eaten was a pear, given that the third fruit eaten is an orange:
Since the third fruit eaten is an orange, we have 7 choices for the third fruit.
Now, we are choosing 2 fruits from the remaining 10 fruits. Out of these 10, there are 3 pears left.
Thus, the number of favorable outcomes in this case is C(3,2) = 3.

Probability = Number of favorable outcomes / Total number of possible outcomes = 3 / 7 ≈ 0.4286

To find the probabilities, we need to know the total number of possible outcomes and the number of favorable outcomes for each scenario.

Total number of possible outcomes:
Since we are choosing 3 fruits out of a total of 11 (4 pears + 7 oranges), the number of possible outcomes can be calculated using combinations. The formula for combinations is C(n, r) = n! / (r!(n-r)!), where n is the total number of items and r is the number of items chosen.

So, the total number of possible outcomes = C(11, 3) = 11! / (3!(11-3)!) = 165.

Now let's calculate the probabilities for each scenario:

(1) Probability that 2 pears and 1 orange are eaten:
We need to calculate the number of favorable outcomes where we choose 2 pears out of 4 (C(4, 2) = 4! / (2!(4-2)!) = 6 ways) and 1 orange out of 7 (C(7, 1) = 7! / (1!(7-1)!) = 7 ways). Then, we multiply these two numbers to get the total number of favorable outcomes.

Number of favorable outcomes = 6 * 7 = 42

Probability = (Number of favorable outcomes) / (Total number of possible outcomes) = 42 / 165 = 14 / 55

Therefore, the probability that 2 pears and 1 orange are eaten is 14/55.

(2) Probability that the third fruit eaten is an orange:
This scenario is independent of the previous scenario. So, we can consider the problem with only 2 fruits remaining. Out of the remaining 2 fruits, 1 is an orange.

Number of favorable outcomes = 1

Probability = (Number of favorable outcomes) / (Total number of possible outcomes) = 1 / 2 = 1/2

Therefore, the probability that the third fruit eaten is an orange is 1/2.

(3) Probability that the first fruit eaten was a pear, given that the third fruit eaten is an orange:
In this scenario, we know that the third fruit eaten is an orange. Now let's calculate the probability that the first fruit eaten was a pear.

Number of favorable outcomes = The number of ways to choose 1 pear out of the 4 pears = C(4, 1) = 4

Probability = (Number of favorable outcomes) / (Total number of possible outcomes) = 4 / 165

Therefore, the probability that the first fruit eaten was a pear, given that the third fruit eaten is an orange, is 4/165.

ppo

4/11 * 3/10 * 7/9 = a

pop
4/11 * 7/10 * 3/9 = b

opp
7/11 * 4/10 * 3/9 = c

add a+b+c
for the answer to (1)
===============================
first is p = 4/11
second is p = 3/10
third is o = 7/9
answer to (2) is 4/11*3/10*7/9

for ( 30), Bayes
try
http://en.wikipedia.org/wiki/Bayes'_theorem

if the third one is an orange
then we really have 10 total
4 pears and 6 oranges to chose from for our first pick
4/10