Question

Mangoes numbered 1 through 18 are placed
in a bag for delivery. Two mangoes are drawn
out of the bag without replacement. Find the
probability such that all the mangoes have
even numbers on them
​

Answer 1

Answer:

Step-by-step explanation:

Originally Answered: A box contains 20 mangoes out of which 4 are not good. If two mangoes are taken out without replacement what is the probability distribution of the number of bad mangoes in the sample?

So there are 20 mangoes, 4 are bad which means 16 are good.

20 × 5 = 100

4 × 5 = 20

16 × 5 = 80

So 20% are bad and 80% are good. If I take out 2 mangoes, I am taking out 2 - 5% chunks out of the total. There are 6 possible combinations of 2 bad mangoes from the bag, while there's 120 possible combinations of 2 good mangoes, and there is 64 combinations of one good and one bad mango. The total number of possibilities is 190.

6/190 = 0.0315789474 which equals 3.15789474% of the possibilities

120/190 = 0.6315789474, which is 63.15789474%

64/190 = 0.3368421053, which is 33.68421053%

So to round it out, 2

Answer 2

The probability of drawing two even mangoes from the bag is 0.235.

Given:

Number of mangoes = 18

Number of mangoes drawn out of the bag = 2

To find:

Probability of two even-numbered mangoes to be drawn.

Solution:

The number of mangoes present is 18. Each of these mangoes is numbered from 1 to 18 as 1, 2, 3,.....18.

The numbers are divided into odd or even.

The odd numbers include: {1, 3, 5, 7, 9, 11, 13, 15, 17} = 9

The even numbers include: {2, 4, 6, 8, 10, 12, 14, 16, 18} = 9

Since we are drawing two mangoes that should be numbered even, the probability of the first mango being even is 9/18 and the second mango is 8/17

Here the mangoes and not replaced and there are 9 even mangoes.

Probability:

9/18 x 8/17

1/2 x 8/17

= 0.235

Therefore the probability of drawing two even mangoes from the bag is 0.235.

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