Question

A life scientist worksheet has the names of 36 animals of these 18 fed on seeds 15 feed on insects and 6 feed on both seed and insects a student randomly selects one animal for a study What is the probability that the chosen animal feeds on seeds or insects or both

Answer 1

11/12

Explanation:

18/36 + 15/36= 33/36 = 11/12

Answer 2

The probability that the chosen animal feeds on seeds or insects or both is:

`(3)/(4)`

Explanation:

Let A denote the event that the animal feed on seeds.

B denote the event that the animal feed on insects

and A∩B denote the event that the animal feed on both seed and insects.

and A∪B denote the event that the animal feeds on seeds or insects or both.

Let P denote the probability of an event.

Now, based on the information from the question we have:

`P(A)=(18)/(36)\\\\P(B)=(15)/(36)\\\\P(A\bigcap B)=(6)/(36)`

Now, we know that:

`P(A\bigcup B)=P(A)+P(B)-P(A\bigcap B)`

Hence, on putting the values we have:

`P(A\bigcup B)=(18)/(36)+(15)/(36)-(6)/(36)\\\\\\P(A\bigcup B)=(18+15-6)/(36)\\\\\\P(A\bigcup B)=(27)/(36)=(3)/(4)`