Question

Three boys agree to meet at midday tomorrow. The probability of each boy remembering
to meet is 3/4 , 4/5 , 5/8
Calculate the probability that
(a) at least two of the boys will remember to meet.
(please write the explanation also)

Answer 1

Task. Three boys agree to meet at midday tomorrow. The probability of each boy remembering to meet is $3/4$, $4/5$, $5/8$. Calculate the probability that at least two of the boys will remember to meet.

Solution. Consider the following 4 events:

$A_{011}$ - “only 1st boy will forget to meet”

$A_{101}$ - “only 2dn boy will forget to meet”

$A_{110}$ - “only 3rd boy will forget to meet”

$A_{111}$ - “only boys will remember to meet”

Thus 0 (resp 1) at poisition $i$ means that $i$-th boy will forget (resp. remember) to meet.

Then these events are collectively exhaustive.

We should compute the probability $p$ that at least two of the boys will remember to meet, i.e. the probability

$p=P(A_{111}\cup A_{110}\cup A_{101}\cup A_{011}).$

Since the events are collectively exhaustive it follows that

$p=P(A_{111}\cup A_{110}\cup A_{101}\cup A_{011})=P(A_{111})+P(A_{110})+P(A_{101})+P(A_{011}).$

Let $A_{i}$ be the event that the $i$-th boy will remember to meet, so

$P(A_{1})=3/4,\qquad P(A_{2})=4/5,\qquad P(A_{3})=5/8.$

Then

$A_{111}=A_{1}\cap A_{2}\cap A_{3},$

$A_{110}=A_{1}\cap A_{2}\cap\overline{A_{3}},$

$A_{101}=A_{1}\cap\overline{A_{2}}\cap A_{3},$

$A_{011}=\overline{A_{1}}\cap A_{2}\cap A_{3}.$

Assume that events $A_{1}$, $A_{2}$ and $A_{3}$ are independent. This means that

$P(A_{1}\cap A_{2}\cap A_{3})=P(A_{1})*P(A_{2})*P(A_{3})=\frac{3}{4}*\frac{4}{5}*\frac{5}{8}=\frac{60}{160},$

$P(A_{1}\cap A_{2}\cap\overline{A_{3}})=P(A_{1})*P(A_{2})*(1-P(A_{3}))=\frac{3}{4}*\frac{4}{5}*\frac{3}{8}=\frac{36}{160}$

$P(A_{1}\cap\overline{A_{2}}\cap A_{3})=P(A_{1})*(1-P(A_{2}))*P(A_{3})=\frac{3}{4}*\frac{1}{5}*\frac{5}{8}=\frac{15}{160}$

$P(\overline{A_{1}}\cap A_{2}\cap A_{3})=(1-P(A_{1}))*P(A_{2})*P(A_{3})=\frac{1}{4}*\frac{4}{5}*\frac{5}{8}=\frac{20}{160}$

Hence the probability $p$ that at least two of the boys will remember to meet is equal to

$p=P(A_{111})+P(A_{110})+P(A_{101})+P(A_{011})=\frac{60}{160}+\frac{36}{160}+\frac{15}{160}+\frac{20}{160}$

$=\frac{60+36+15+20}{160}=\frac{131}{160}.$

Answer. $p=\frac{131}{160}$.