1 If one of the parents suffer from the disease X then proba

1) If one of the parents suffer from the disease X, then probability for offsprings to have disease X is 0.5. Calculate probability, that in a family with total 2 childrens, all children are healthy; two children are diseased.

2) There are 9 rats in a cage, 3 of them ar marked. Rats are drawn from the cage one by one. What is the probability, that second rat will be markd, given first rat was marked. How this probability will changes, given first rat was not marked?

Solution

(2)

The probability of getting first rat marked is

P(first rat marked) = 3/ 9

And probability of getting first and second rat marked will be

P(second rat marked and first rat marked) = (3/9)*(2/8)

So the probability, that second rat will be markd, given first rat was marked is

P(second rat marked | first rat marked) = P(second rat marked and first rat marked) / P(first rat marked) = 2 / 8 =1/4

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The probability of getting first rat not marked is

P(first rat not marked) = 6 / 9

And probability of getting first rat not marked and second rat marked will be

P(second rat marked and first rat not marked) = (6 / 9)*(3 / 8)

So the probability, that second rat will be markd, given first rat was not marked is

P(second rat marked | first rat not marked) = P(second rat marked and first rat not marked) / P(first rat not marked) = 3 / 8