Question

There are 15 oranges in a basket. 4 oranges are rotten. Danial takes two oranges at random from the basket without replacement. Find the probability that (a) both oranges are rotten. (b) only one of the oranges is rotten.

who can help me pls​

Answer 1

Part (a)

• 4 rotten
• 15 total

4/15 is the probability of getting a rotten orange

After one is taken and not put back, there are

• 4-1 = 3 rotten
• 15-1 = 14 total

meaning 3/14 is the probability of a second rotten orange.

Two rotten in a row has the probability (4/15)*(3/14) = 2/35. I skipped a few steps so let me know if you need to see in more detail how I got to that result.

Answer: 2/35

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Part (b)

We want to calculate the probability exactly one of the two oranges selected is rotten.

There are two possible cases here:

• Case 1) The first orange is rotten and the second is not rotten.
• Case 2) The first orange is not rotten and the second is rotten.

Let's determine the probability for case 1.

4/15 was mentioned earlier in part (a) as the probability of a rotten orange. There are 15-4 = 11 non-rotten oranges and 15-1 = 14 remaining since we aren't replacing the first selection. That gives 11/14 as the probability the 2nd selection is fresh.

We get (4/15)*(11/14) = 22/105 as the probability for case 1 to happen.

Through similar calculations, the probability for case 2 is (11/15)*(4/14) = 22/105

Cases 1 and 2 are mutually exclusive. Both cannot happen simultaneously. There is no overlap. This fact allows us to add the results to get:

22/105 + 22/105 = 44/105

That fraction cannot be reduced.

Answer: 44/105