Probability

Probability Level 3

IF $A$ and $B$ are two independent events such that $P (A^c \cap B ) =\dfrac2{15}$ , and $P (A \cap B^c) = \dfrac16$ , find $P(A)$ and $P(B)$ .

P(A)=\frac{1}{5},\frac{5}{6}, P(B)= \frac{4}{5},\frac{1}{6}

P(A)=\frac{1}{5},\frac{7}{11}, P(B)= \frac{4}{5},\frac{4}{11}

P(A)=\frac{3}{8},\frac{2}{7}, P(B)= \frac{5}{8},\frac{4}{7}

None of the above

P(A)=\frac{2}{3},\frac{5}{6}, P(B)= \frac{1}{3},\frac{1}{6}

1 solution

Tom Engelsman
Feb 15, 2016

Knowing that for any event X => P(X) + P(X') = 1, let us write:

P(A' and B) = P(A')P(B) = [1 - P(A)]*P(B) = P(B) - P(A)P(B) = 2/15 (i)

P(A and B') = P(A)P(B') = P(A)*[1 - P(B)] = P(A) - P(A)P(B) = 1/6 (ii).

Subtracting (i) from (ii) produces P(A) - P(B) = 1/30, or P(A) = P(B) + 1/30 (iii). Substitution of (iii) into (ii) yields:

P(A) [1 - P(B)] = 1/6 => [1/30 + P(B)][1 - P(B)] = 1/6 => 1/30 + (29/30) P(B) - P(B)^2 = 1/6 (iv). Solving the quadratic equation in (iv) for P(B) produces P(B) = 1/6, 4/5, and plugging these values into (iii) yields P(A) = 1/5, 5/6 respectively.

Hence the answer is P(A) = 1/5, 5/6 and P(B) = 1/6, 4/5.

In the "correct" answer of the problem is written P(B)=4/5, 1/6! This is not 1/6, 4/5 with which I agree. This is why I answered None etc.

Wim Kerstens - 5 years, 3 months ago

I agree the answer choices are worded a little confusingly, Wim! It would've made more sense to group P(A) & P(B) as ordered-pairs instead. Kudos for still trying everything out :)

tom engelsman - 5 years, 3 months ago

If (A and B) are independent events then (A' and B) and (A and B') must be dependent events. Visualize this using Venn diagram , so the above formula used in ans is wrong rather, this formula should be used --> P(A' and B) = P(A')P(B|A') or P(A and B')=P(B)P(A'|B). Please let me know if I'm wrong.

GAURAV YADAV - 1 year, 5 months ago

Probability

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