NEWEST Probability (P) Exam of Society of Actuaries (SOA) | ULTIMATE EXAM WITH CORRECT ANSWERS AND RATIONALES FOR CERTIFICATION SUCCESS

NEWEST Probability (P) Exam of Society of
Actuaries (SOA) | ULTIMATE EXAM WITH
CORRECT ANSWERS AND RATIONALES
FOR CERTIFICATION SUCCESS
1. A fair six-sided die is rolled twice. Let X be the sum
of the two rolls. Calculate the probability that X is odd
given that the first roll is odd.
A) 1/4
B) 1/3
C) 1/2
D) 2/3
E) 3/4
Correct answer: C
Rationale: First roll odd (1,3,5). Second roll odd gives
even sum; second roll even gives odd sum.
P(even|first odd)=1/2 independent.


2. An insurance company determines that 80% of
drivers have no accidents in a year, 15% have exactly
one accident, and 5% have two or more accidents.
For those with no accidents, the probability of a claim
is 0.1; for one accident, 0.8; for two or more, 0.9.

,Calculate the probability that a randomly selected
driver files a claim.
A) 0.15
B) 0.20
C) 0.25
D) 0.30
E) 0.35
Correct answer: C
Rationale: Law of total probability: 0.8×0.1 + 0.15×0.8
+ 0.05×0.9 = 0.08 + 0.12 + 0.045 = 0.245 ≈ 0.25.


3. A random variable X has probability density
function f(x) = kx^2 for 0 ≤ x ≤ 3, and 0 otherwise. Find
the value of k.
A) 1/27
B) 1/18
C) 1/9
D) 1/6
E) 1/3
Correct answer: C
Rationale: ∫₀³ kx² dx = k(27/3)=9k=1 ⇒ k=1/9.

,4. For a continuous random variable X with
cumulative distribution function F(x)=1−e^{−x/2} for
x≥0, find the median of X.
A) 2 ln 2
B) ln 2
C) 2
D) 1
E) 0.5
Correct answer: A
Rationale: Set F(m)=0.5 ⇒ 1−e^{−m/2}=0.5 ⇒
e^{−m/2}=0.5 ⇒ −m/2=ln(0.5) ⇒ m=2 ln 2.


5. A box contains 5 red and 7 blue balls. Two balls are
drawn without replacement. What is the probability
that the second ball is red given that the first ball is
blue?
A) 5/12
B) 5/11
C) 7/12
D) 7/11
E) 1/2
Correct answer: B

, Rationale: After one blue removed, 5 red and 6 blue
remain. P(second red|first blue)=5/(5+6)=5/11.


6. Let X and Y be independent Poisson random
variables with means 2 and 3 respectively. Find
P(X+Y=1).
A) e^{−5}×5
B) e^{−5}×6
C) e^{−5}×7
D) e^{−5}×8
E) e^{−5}×9
Correct answer: A
Rationale: X+Y ~ Poisson(5). P(S=1)=e^{−5}×5^1/1!
=5e^{−5}.


7. The moment generating function of a random
variable X is M_X(t) = (1−2t)^{−3} for t<1/2. Find E[X].
A) 2
B) 3
C) 4
D) 5
E) 6