Free SOA Exam P (Probability) Probability Fundamentals Practice Questions

$P(A \cap B) = P(B \mid A) \times$ P(A) = 0.4 $\times$0.7 = 0.28.

MISSISSIPPI has 11 letters: M(1), I(4), S(4), P(2).

The number of distinct arrangements is:

$\frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!} = \frac{39916800}{1 \times 24 \times 24 \times 2} = \frac{39916800}{1152} = 34650$

Distractor analysis:
- 39916800: Computed 11! without dividing by repeated letters.
- 69300: Divided by $4! \times 4!$ only, forgetting the 2! for P: $39916800/576 = 69300$.
- 11!: Wrote the factorial symbol without evaluating or dividing.
- 2520: Divided by too many repeated counts, e.g., $\frac{11!}{4!\cdot4!\cdot2!\cdot1!\cdot \text{extra}}$.

The answer is $34650$.

P(A∪B') = 1 - P(B) + P(A∩B) = 0.9, so P(A∩B) = P(B) - 0.1. Substituting into P(A∪B) = P(A) + P(B) - P(A∩B) gives P(A) + 0.1 = 0.7, so P(A) = 0.6.