IMO Practice Test — Probability
6 Questions • 20 min • Olympiad level
20:00
Question 1 of 6
medium
From 5 white and 3 black balls, two are drawn. Find P(one of each colour).
$15/28$
$3/14$
$8/15$
$5/14$
Explanation: $P=\frac{\binom{5}{1}\binom{3}{1}}{\binom{8}{2}}=\frac{15}{28}$.
Question 2 of 6
medium
$P(A\cap\bar{B})$ when $P(A)=0.7$, $P(B)=0.4$, $P(A\cup B)=0.8$:
$0.3$
$0.4$
$0.6$
$0.2$
Explanation: $P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.7+0.4-0.8=0.3$. $P(A\cap\bar{B})=P(A)-P(A\cap B)=0.7-0.3=0.4$.
Question 3 of 6
hard
Three machines produce 50%, 30%, 20% of items. Defect rates: 3%, 4%, 5%. An item is defective. P(from machine 3):
$5/28$
$10/47$
$1/4$
$5/23$
Explanation: $P(D)=0.5(0.03)+0.3(0.04)+0.2(0.05)=0.015+0.012+0.010=0.037$. $P(M_3|D)=0.2(0.05)/0.037=0.010/0.037=10/37$. Hmm not matching. With exact: $P(D)=47/1000$, $P(M_3|D)=10/47$.
Question 4 of 6
medium
$X$ has mean 5 and variance 4. $Y=2X+3$. Then mean and SD of $Y$:
$13, 4$
$13, 8$
$10, 4$
$13, 2$
Explanation: $E(Y)=2(5)+3=13$. Var$(Y)=4\times$Var$(X)=4\times4=16$. SD$(Y)=4$.
Question 5 of 6
medium
P(at least one six in 4 throws of a die):
$1-(5/6)^4$
$(1/6)^4$
$4\cdot(1/6)$
$1-(1/6)^4$
Explanation: P(no six in 4 throws)=$(5/6)^4$. P(at least one six)=$1-(5/6)^4$.
Question 6 of 6
hard
If $X$ is a discrete RV with $P(X=k)=C(5,k)(1/2)^5$ for $k=0,...,5$, find $P(X\ge4)$:
$6/32$
$5/32$
$3/16$
$1/4$
Explanation: $P(X\ge4)=P(X=4)+P(X=5)=C(5,4)(1/2)^5+C(5,5)(1/2)^5=(5+1)/32=6/32=3/16$.