Question

Three boys agree to meet at midday tomorrow. The probability of each boy remembering

to meet is 3/4 , 4/5 , 5/8

Answer 1

Consider the following 4 events:

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

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

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

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

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

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

Since the events are collectively exhaustive it follows that

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

$$=P(A_{011})+P( A_{101})+P( A_{110})+P( A_{111})$$

$$A_{011}=A_1^C\cap A_2\cap A_3$$

$$A_{101}=A_1\cap A_2^C\cap A_3$$

$$A_{110}=A_1\cap A_2\cap A_3^C$$

$$A_{111}=A_1\cap A_2\cap A_3$$

Given $P(A_1)=\dfrac{3}{4}, P(A_2)=\dfrac{4}{5}, P(A_3)=\dfrac{5}{8}.$

Assume that events $A_1, A_2$ and $A_3$ are independent. This means that

$$P(A_{011})=P(A_1^C)P(A_2)P(A_2)=$$

$$=(1-\dfrac{3}{4})(\dfrac{4}{5})(\dfrac{5}{8})=\dfrac{1}{8}$$

$$P(A_{101})=P(A_1)P(A_2^C)P(A_2)=$$

$$=\dfrac{3}{4}(1-\dfrac{4}{5})(\dfrac{5}{8})=\dfrac{3}{32}$$

$$P(A_{110})=P(A_1)P(A_2)P(A_2^C)=$$

$$=\dfrac{3}{4}(\dfrac{4}{5})(1-\dfrac{5}{8})=\dfrac{9}{40}$$

$$P(A_{111})=P(A_1)P(A_2)P(A_2)=$$

$$=\dfrac{3}{4}(\dfrac{4}{5})(\dfrac{5}{8})=\dfrac{3}{8}$$

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

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

$$=\dfrac{1}{8}+\dfrac{3}{32}+\dfrac{9}{40}+\dfrac{3}{8}=\dfrac{131}{160}=0.81875$$