# Thread: Sample probability question from an actuarial test

1. ## Sample probability question from an actuarial test

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A survey of a group’s viewing habits over the last year revealed the following
information:

(i) 28% watched gymnastics
(ii) 29% watched baseball
(iii) 19% watched soccer
(iv) 14% watched gymnastics and baseball
(v) 12% watched baseball and soccer
(vi) 10% watched gymnastics and soccer
(vii) 8% watched all three sports.

Calculate the percentage of the group that watched none of the three sports
during the last year.
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The way I approached the question is like this.

Let
P(i) = 28%
P(ii) = 29%
P(iii) = 19%
P(iv) = 14%
P(v) = 12%
P(vi) = 10%
P(vii) = 8%

Obviously, the answer isn't simply 100% - P(i) - P(ii) - P(iii) because there are bunch of "overlaps" going on. To compensate for these "overlaps", you need to add back P(iv), P(v), and P(vi). But this would be adding back too much. So, you need to take away a certain multiple of P(vii). I thought you needed to take away 3 * P(vii), but the answer says take away 1 * P(vii).

If you followed what I said, I thought the answer would be

100% - P(i) - P(ii) - P(iii) + P(iv) + P(v) + P(vi) - P(vii) - P(vii) - P(vii) = 36

100% - P(i) - P(ii) - P(iii) + P(iv) + P(v) + P(vi) - P(vii) = 52

Can someone explain this?

(I wish I can draw a picture on here so I can explain what I'm thinking/visualizing. When I did this, I imagined 3 colored circles overlapping and the areas I would take away and add back to get an area where there wouldn't be any overlaps. When I visualized the problem in this way I thought the answer I came up with was correct, but it wasn't)

2. Hello, doolindalton!

A survey of a group’s viewing habits over the last year:

(a) 28% watched gymnastics
(b) 29% watched baseball
(c) 19% watched soccer
(d) 14% watched gymnastics and baseball
(e) 12% watched baseball and soccer
(f) 10% watched gymnastics and soccer
(g) 8% watched all three sports.

Find the percentage that watched none of the three sports.
We are given:

. . $\begin{array}{ccc} P(G) &=& 28\% \\ P(B) &=& 29\% \\ P(S) &=& 19\% \\ P(G \cap B) &=& 14\% \\ P(B \cap S) &=& 12\% \\ P(G \cap S) &=& 10\% \\ P(G \cap B \cap S) &=& 8\% \end{array}$

We are expected to know this formula:

. . $P(G \,\cup\,B \,\cup \,S) \;=\;P(G) + P(B) + P(S) - P(G \,\cap \,B ) - P(B \,\cap\, S)\; -$ $P(G \,\cap\, S) + P(G \,\cap\, B \,\cap S)$

We have: . $P(G \,\cup\, B \,\cup \,S) \;=\;28 + 29 + 19 - 14 - 12 - 10 + 8 \;=\;48$

. . That is, 48% watched at least one of the sports.

Therefore: . $100\% - 48\% \;=\;{\color{blue}52\%}$ watched none of the sports.

3. Originally Posted by Soroban
Hello, doolindalton!

We are expected to know this formula:

. . $P(G \,\cup\,B \,\cup \,S) \;=\;P(G) + P(B) + P(S) - P(G \,\cap \,B ) - P(B \,\cap\, S)\; -$ $P(G \,\cap\, S) + P(G \,\cap\, B \,\cap S)$

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Well, as I stated in the original question, I know that is the formula that gives the correct answer, but my question is why do you not need to add that last piece 3 times?