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Probability that at least 2 of 3 approve =

(1) — (Probability of either: only 1 approves OR none approve)

P(A) = .7
P(B) = .5
P(C) = .4

Case 1: probability only 1 approves

(.7) (.5) (.6)
+
(.3) (.5) (.6)
+
(.3) (.5) (.4) =

(.21) + (.09) + (.06) =

.36

Case 2: probability NONE approve

(.3) (.5) (.6) =

.09

Probability of Case 1 OR Case 2 = (.36) + (.09) = .45

1 - (.45) = .55

*answer*
.55
D

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Probability that at least 2 of 3 approve =

(1) — (Probability of either: only 1 approves OR none approve)

P(A) = .7
P(B) = .5
P(C) = .4

Case 1: probability only 1 approves

(.7) (.5) (.6)
+
(.3) (.5) (.6)
+
(.3) (.5) (.4) =

(.21) + (.09) + (.06) =

.36

Case 2: probability NONE approve

(.3) (.5) (.6) =

.09

Probability of Case 1 OR Case 2 = (.36) + (.09) = .45

1 - (.45) = .55

*answer*
.55
D

Posted from my mobile device
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I wonder if there's a more effective way to solve this other than just doing it the calculation heavy way.
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I dont think you can do it without those heavy calculations but since there is a 0.5 in every calculation you could save some time with neglecting the 0.5 in the first place and just coming back to it in the end so you can come up to the answer in under a minute:

2 approve:

7*4 = 28
7*6 = 42
3*4 = 12

3 approve:

7*4 = 28

All added up = 110

Times 0.5 = 55

/100 = 0.55
KevoK
I wonder if there's a more effective way to solve this other than just doing it the calculation heavy way.
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Hi KevoK,

Great question — there IS a slightly slicker way to organize this. Instead of listing all 4 winning scenarios separately, try conditioning on one VP.

Pick the easiest VP to work with. Here, Baker is nice because P(B) = 0.5 and P(B') = 0.5.

Split into two worlds:

World 1: Baker approves (probability 0.5)
Johnson now only needs 1 more approval from Adams or Corfu. It's easier to find the complement — neither approves: 0.3 × 0.6 = 0.18. So P(at least 1 of them approves) = 1 - 0.18 = 0.82.

World 2: Baker doesn't approve (probability 0.5)
Johnson now needs BOTH Adams AND Corfu: 0.7 × 0.4 = 0.28.

Combine:
0.5 × 0.82 + 0.5 × 0.28 = 0.41 + 0.14 = 0.55

Answer: D

Why this is faster: instead of computing 4 separate three-way products and adding them up, you only need 2 quick sub-calculations. The trick is to condition on one person, which reduces the 'at least 2 out of 3' problem into simpler 'at least 1 out of 2' and 'both out of 2' problems.

General principle: When a probability problem feels calculation-heavy, try splitting on one variable to break it into simpler sub-problems. This 'divide and conquer' approach often cuts the work roughly in half.
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