1. ## Basic Probabilty

Of 500 students , 200 participating in math activity , 250 participating in science activity and 50 participated in both activities , Find that a randomly selected student.
a) Will be a participant in at least one of the two activities.
b) Will not be a participant in either activity.
c) Will be a participant in exactly one activity.

a) 0.8
b) 0.3
c) 0.5

2. For the first one, yes.

But for the other two... can I ask you how you got those answers?

3. First , Thank you
Second , here is it
P(M)=0.4
P(S)=0.5
P(M AND S)=0.1

a)
P(M OR S)= P(M)+P(S)-P(M AND S)

b)
P(M')=0.6
P(S')=0.5
P(M' AND S') = P(M') . P(S')

c)
P(S' AND M) OR P(S AND M') = P(M) . P(S') +[P(M') . P(S)]

4. b). Ah, there is your mistake. The probability that the student will not be a participant in either activities is:

$P(M \cup S)' = 1 - P(M \cup S) = 1 - 0.8 = 0.2$

Drawing a Venn Diagram might be easier to follow.

c). Here also, a Venn Diagram would show you the right answer. You didn't remove the set $M \cap S$

Let me show you the Venn Diagram:

You will immediately see that those who will not be participant in either activities is 100 over 500.

And for c), it becomes (150 + 200)/500 = 0.7

5. Thank a lot
But I'm still confused for part c
who will not be participant in either activities is 100 over 500.
From where you got 100 student will not be participant??

6. They are those which aren't in any of the sets.

You know that 50 students perform both, hence the 50 in the intersection of the sets.

You know a total of 200 perform in math, hence, 150 performed only math.
You know a total of 250 perform in science, hence, 200 performed only science.

Total up to now is 150 + 50 + 200 = 400

which means, 100 of them don't do either.

The probability then becomes obvious.

7. aha got it thank you