Answer to Question #156446 in Statistics and Probability for Nathan Smith

Question

Jamal has waffles for breakfast every Sunday morning. He tops them with blueberry syrup or maple syrup, but sometimes he uses both or neither of the syrups. 

On one Sunday:

Let X be the event ‘Jamal uses blueberry syrup’

Let Y be the event ‘Jamal uses maple syrup’

It is given that

P(X) = 13/30

P(X | Y) = 1/4

P(X | Y ') = 9/14

Find the probability that Jamal uses both blueberry and maple syrup

Answer 1

Given that $P(X)=\frac{13}{30}$ ,$P(X/Y)=\frac{1}{4},P(X/Y')=\frac{9}{14}$

We have to find $P(X\cap Y).$

As $P(X/Y)=\frac{1}{4}$

$\implies \frac{P(X\cap Y)}{P(Y)}=\frac {1}{4}$

$\implies P(X \cap Y)=\frac {1}{4}.P(Y)$ .......(1)

Again, $P(X/Y')=\frac {9}{14}$

$\implies \frac { P(X\cap Y')}{P(Y')}=\frac{9}{14}$

$\implies \frac {P(X\cup Y)-P(Y)}{1-P(Y)}=\frac {9}{14}$ [ as $P(X\cup Y)=P(X\cap Y')+P(Y),$ where $Y$ and $X\cap Y'$ are disjoint set ]

$\implies 14.P(X\cup Y)=5.P(Y)+9$

$\implies 14.[P(X)+P(Y)-P(X\cap Y)]=5.P(Y)+9$

$\implies 14.[P(X)+P(Y)-\frac{1}{4}P( Y)]=5.P(Y)+9$

$\implies \frac{11}{2}.P(Y)=9-14.P(X)$

$\implies \frac{11}{2}.P(Y)=9-14.(\frac{13}{30})$

$\implies \frac{11}{2}.P(Y)=9-\frac{91}{15}$

$\implies P(Y)=\frac{8}{15}$

Therefore from (1), we have $P(X \cap Y)=\frac {1}{4}.\frac{8}{15}=\frac{2}{15}$

$\therefore$ The required probability that Jamal uses both blueberry and maple syrup is $=\frac{2}{15}$