Question

In a basket of rambutans and bananas ,70% of rambutans and 60% of the bananas are ripe.If 40% of the fruits in the basket re rambutans.find the probability of the fruits which are ripe.Find probability of rambutans given the fruits are ripe?

Answer 1

We have a `64%` chance of drawing a ripe fruit. `43.8%` of the ripe fruit are rambutans.

Explanation:

Lets write what we know about the population of fruits.

Fraction of rambutans that are ripe: `r_r=0.7`
Fraction of bananas that are ripe: `b_r=0.6`
Fraction of rambutans in total: `r_t = 0.4`

Therefore we also know

Fraction of bananas in total: `b_t=0.6` (assuming that there are no other fruits that we are not told about, `r_t+b_t = 1`). If we randomly select a fruit from the basket we can calculate the probability that the fruit is ripe from:

In words, the probability that the fruit is a rambutan AND this rambutan is ripe, PLUS the probability the fruit is a banana AND the banana is ripe:

`P_r=r_t*r_r+b_t*b_r = 0.4*0.7+0.6*0.6 = 0.28+0.36 = 0.64`

In words, we have a `64%` chance of drawing a ripe fruit.

Now we want to know, of all the ripe fruit, what fraction are rambutans? We can use the previous formula to determine this since it represents the fraction of fruit that are ripe, we can ask what fraction of those are rambutans.

`P_(r,r)=(r_t*r_r)/P_r=(0.4*0.7)/0.64=0.4375`

In words, `43.8%` of the ripe fruit are rambutans.

We can check this with a specific example by taking a population of 100 fruits.

From this we see that 64 out of 100 fruit are ripe, therefore we have a `64%` probability of drawing a ripe fruit. We can also see that 28 out of the 64 ripe fruit are rambutans, meaning that we have a `28//64` or `43.8%` probability of drawing a rambutan from the population of ripe fruit.