Question

Two numbers are randomly selected and multiplied. Consider two events $E_1$ and $E_2$ defined as
$E_1$ : Their product is divisible by 5
$E_2$ : Unit’s places in their product is 5
Which of the following statement is/are correct?

• A. $E_1$ is twice as likely to occur as $E_2$
• B. $E_1$ and $E_2$ are disjoint
• C. $P\left(E_2/E_1\right)=\frac{1}{4}$
• D. $P\left(E_1/E_2\right)=1$

Answer 1

Answer: (C), (D)

The correct options are
C $P\left(E_2/E_1\right)=\frac{1}{4}$
D $P\left(E_1/E_2\right)=1$

$P\left(E_1\right)=1-P(\text{unit's place in both is}1,2,3,4,6,7,8,9$
$P(E_1=0\text{or}5)=1-{\left(\frac{4}{5}\right)}^2=\frac{9}{25}$
$P(E_2:5)=P\left(13579\right)-P\left(1379\right)$
$=\frac{1}{4}-\frac{4}{25}$
$=\frac{9}{100}$
$\frac{P\left(E_2\right)}{P\left(E_1\right)}=\frac{9}{100}\times \frac{25}{4}=\frac{1}{4}$
$P\left(E_1\right)=4P\left(E_2\right)$
$P\left(E_2/E_1\right)=\frac{P(E_2\cap E_1)}{P\left(E_1\right)}=\frac{P\left(E_2\right)}{P\left(E_1\right)}=\frac{1}{4}$
$P\left(E_1/E_2\right)=\frac{P(E_1\cap E_2)}{P\left(E_2\right)}=\frac{P\left(E_2\right)}{P\left(E_2\right)}=1$