Question

The probability that a student $x$ participates in a competition is $0.4$, while that for student $y$ is $0.5$. The probability that $x$ participates in competition, given that $y$ also participates is $0.7$. Then

• A. The probability that both $x$ and $y$ participates in that competition simultaneously is $0.35$
• B. The probability that $y$ participates in that competition given that $x$ also participates is $\frac{7}{8}$
• C. The probability that atleast one of them participates in that competition is $0.55$
• D. The probability that exactly one of them participates in that competition is $0.20$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The probability that exactly one of them participates in that competition is $0.20$
B The probability that both $x$ and $y$ participates in that competition simultaneously is $0.35$
C The probability that $y$ participates in that competition given that $x$ also participates is $\frac{7}{8}$
D The probability that atleast one of them participates in that competition is $0.55$

$P\left(x\right)=0\cdot 4$
$P\left(y\right)=0\cdot 5$
$P\left(x|y\right)=0\cdot 7$
(A) $P(x\cap y)=P\left(y\right)P\left(x|y\right)=(0\cdot 5)(0\cdot 7)=0\cdot 35$

(B) $P\left(y|x\right)=\frac{P(x\cap y)}{P\left(x\right)}=\frac{0\cdot 35}{0\cdot 4}=\frac{35}{40}=\frac{7}{8}$

(C) $P(x\cup y)=P\left(x\right)+P\left(y\right)-P(x\cap y)$

$=0\cdot 4+0\cdot 5-0\cdot 35=0\cdot 90-0\cdot 35=0\cdot 55$

(D) $P(x\cap \overline{y})+P(\overline{x}\cap y)=P\left(x\right)-P(x\cap y)+P\left(y\right)-P(x\cap y)$
$=0\cdot 4+0\cdot 5-2(0\cdot 35)$
$=0\cdot 9-0\cdot 7=0\cdot 2$