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

Answer:

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

                $\dfrac{3}{4}$

Step-by-step 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:

$P(A)=\dfrac{18}{36}\\\\P(B)=\dfrac{15}{36}\\\\P(A\bigcap B)=\dfrac{6}{36}$

Now, we know that:

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

Hence, on putting the values we have:

$P(A\bigcup B)=\dfrac{18}{36}+\dfrac{15}{36}-\dfrac{6}{36}\\\\\\P(A\bigcup B)=\dfrac{18+15-6}{36}\\\\\\P(A\bigcup B)=\dfrac{27}{36}=\dfrac{3}{4}$

Answer 2

Answer:

11/12

Step-by-step explanation:

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