Question

Jane has a bag with 5 oranges, 4 grapefruits, and 3 apples. What is the probability that she will NOT randomly take out two oranges in a row? What is the probability that she will NOT randomly take out two grapefruits in a row? What is the probability that she will NOT randomly take out two apples in a row? What is the probability that she will NOT take out an orange, then an apple, and finally a grapefruit in that order?

Once she has removed a fruit, she will put it back in before removing a second fruit.***

Answer 1

To find the probability of not randomly taking out two oranges in a row, we need to consider the total number of fruits in the bag before and after each draw. Since Jane puts the fruit back in before removing a second fruit, the probability of not drawing an orange twice in a row would be the same as the probability of drawing a non-orange fruit after the first draw.

Similarly, we can calculate the probability of not randomly taking out two grapefruits in a row and two apples in a row using the same logic.

To find the probability of not taking out an orange, then an apple, and finally a grapefruit in that order, we multiply the probabilities of each individual event happening in that order.

Let's calculate the probabilities step by step:

1. Probability of not randomly taking out two oranges in a row:
Total fruits in the bag = 5 oranges + 4 grapefruits + 3 apples = 12
Probability of drawing a non-orange fruit after the first draw = (12 - 5) / 12

2. Probability of not randomly taking out two grapefruits in a row:
Total fruits in the bag = 5 oranges + 4 grapefruits + 3 apples = 12
Probability of drawing a non-grapefruit fruit after the first draw = (12 - 4) / 12

3. Probability of not randomly taking out two apples in a row:
Total fruits in the bag = 5 oranges + 4 grapefruits + 3 apples = 12
Probability of drawing a non-apple fruit after the first draw = (12 - 3) / 12

4. Probability of not taking out an orange, then an apple, and finally a grapefruit in that order:
Probability of not taking out an orange first = (12 - 5) / 12
Probability of not taking out an apple second = (12 - 3) / 12
Probability of not taking out a grapefruit third = (12 - 4) / 12
Multiply the probabilities together: [(12 - 5) / 12] * [(12 - 3) / 12] * [(12 - 4) / 12]

Please note that these calculations assume that each fruit has an equal probability of being drawn and that the fruits are replaced in the bag after each draw.

Answer 2

First let’s take into account that there are 12 fruits total at the start, three of those being apples. The chance that he gets an apple first is 3 in 12, or 1 in 4. Then, an apple is taken out, and the count of fruits in the bag is brought down to two. The chances that he gets another apple would be 2 in 11.

To get the probability that he does get two apples in a row, we multiply 3/12 by 2/11 to get 6/132 or a 1/22 chance.

To get the probability that he doesn’t get two apples in a row, we must understand that the chance of him not getting two apples in a row is the opposite of the chance that he does get two in a row. So the probability that he will not randomly take out two apples in a row is 1 - 1/22 = 21/22

I think this is the answer