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3 Students, $A, B, C$ answer a question (independently). The probability of $A$ getting the correct answer is $0.9$, probability for $B$ is $0.7$ and the probability for $C$ is $0.4$ Q: Find the probability exactly two students get the question correct.

We can have $AB, BC, AC$ as those who get it correct. The question is, what about $ABC$?

I have

$P(\text{Exactly two get it right}) = P(AB \cup BC \cup AC) = P(AB) + P(AC) + P(BC) - P(AB \cap AC) - P(AB \cap BC) - P(BC \cap AC) + P(AB \cap AC \cap BC) $

But isn't that just

$$= P(AB) + P(AC) + P(BC) - 2 \cdot P(AB \cap AC \cap BC) $$

$$= (0.9*0.7) + (0.9*0.4) + (0.7*0.4) - 2(???)$$

Can we assume $AB, AC, BC$ are independent? What do I do?

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    $\begingroup$ You can simplify it. You want two of them (for example A,B) to be correct and the third (i.e C) to be wrong. Probability of this case is just $P(A\cap B\cap \bar{C})=P(A)P(B)P(\bar{C})$. Think which cases you have. $\endgroup$ – Galc127 Jan 24 '17 at 14:01
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    $\begingroup$ $AB \cup BC \cup AC$ is not the event that exactly two students get it right $\endgroup$ – Canardini Jan 24 '17 at 14:04
  • $\begingroup$ @Galc127, so it would 3 times that for the complete probability (getting all cases)? $\endgroup$ – Amad27 Jan 24 '17 at 14:15
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P(A right, B right, C wrong): $0.9 \times 0.7 \times 0.6=0.378$

P(A right, B wrong, C right): $0.9 \times 0.3 \times 0.4=0.108$

P(A wrong, B right, C right): $0.1 \times 0.7\times 0.4=0.028$

Required probability: $0.378+0.108+0.028=0.514$

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Split it into disjoint events, and then add up their probabilities:

  • $P({A}\cap{B}\cap\overline{C})=(0.9)\cdot(0.7)\cdot(1-0.4)=0.378$
  • $P({A}\cap\overline{B}\cap{C})=(0.9)\cdot(1-0.7)\cdot(0.4)=0.108$
  • $P(\overline{A}\cap{B}\cap{C})=(1-0.9)\cdot(0.7)\cdot(0.4)=0.028$

The probability is therefore $0.378+0.108+0.028=0.514$.

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Case 1 - Probability of not solved by A

$P(\overline A) = 1 - P(A)$

= 1 - 0.9 = 0.1

Case 2 - Probability of not solved by B

$P(\overline B) = 1 - P(B)$

= 1 - 0.7 = 0.3

Case 3 - Probability of not solved by C

$P(\overline C) = 1 - P(C)$

= 1 - 0.4 = 0.6

Probability of exactly two solve

$ = P(\overline A).P(B).P(C) + P(A).P(\overline B).P(C) + P(A).P(B).P(\overline C)$

Putting the values,

= 0.1 × 0.7 × 0.4 + 0.9 × 0.3 × 0.4 + 0.9 × 0.7 × 0.6

= 0.028 + 0.108 + 0.378 = 0.514

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