Answer to Question #256774 in Statistics and Probability for John

Question

The probabilities that Dorothy will join the Mathematics, English and Judo clubs are 0.4,

0.45 and 0.55, respectively. The probability that she will join both the Mathematics and

English clubs is 0.2. If Dorothy has decided to join the Mathematics club, the probability that

she will not join the Judo club is 0.6. If Dorothy has decided to join both the Mathematics

and English clubs, the probability that she will not join the Judo club is 0.7. On the other

hand, if Dorothy has decided to join the Judo club, the probabilities that she will join the

English club become 0.4.

(a)

What is the probability that Dorothy will join either the Mathematics club or the English club?

(b)

What is the probability that Dorothy will join all the three clubs?

(c)

What is the probability that Dorothy will join only one club?

Answer 1

Let M be the event that Dorothy joins Mathematics club, E be the event that she joins English club and J be the event that she joins Judo club.

First, let us define the following probabilities from the above information.

$p(M)=0.4$

$p(E)=0.45$

$p(J)=0.5$

The probability that she will join both the Mathematics and English clubs is denoted by

$p(M\cap E)=0.2$ .

If Dorothy has decided to join the Mathematics club, the probability that she will not join the Judo club is a conditional probability given as, $p(J'|M)=0.6$.

This conditional probability implies that the conditional probability that she joins Judo club if she has decided to join Mathematics club is given as,

$p(J|M)=1-P(J'|M)=1-0.6=0.4.$

If Dorothy has decided to join both the Mathematics and English clubs, the probability that she will not join the Judo club is the conditional probability given as, $p(J'|M\cap E)=0.7$

a.

The probability that Dorothy will join either the Mathematics club or the English club is denoted by $p(M\cup E)$ and is given by,

$P(M\cup E)=p(M)+p(E)-p(M\cap E)=0.4+0.45-0.2=0.65$

Thus, the probability that Dorothy will join either Mathematics or English club is 0.65.

b.

The probability that Dorothy will join all the three clubs is the intersection of the 3 events and it is given as, $p(M\cap E\cap J).$

We are given that, $p(J'|M\cap E)=0.7$. From this, we can determine $p(J|M\cap E)=1-P(J'|M\cap E)=1-0.7=0.3$

From definition of conditional probability, we have,

$p(J|M\cap E)=p(J\cap M\cap E)/P(M\cap E)$.

This implies that,

$p(J\cap M\cap E)=p(J|M\cap E)*p(M\cap E)=0.3*0.2=0.06$

Therefore, the probability that Dorothy will join all the three clubs is 0.06.

c.

The probability that Dorothy will join only one club is given as,

$p(exactly \space one\space event )=p(M)+p(E)+p(J)-2(p(M\cap E)+p(M\cap J)+p(E\cap J))+3p(M\cap E\cap J)$

Since we have,

$p(M)=0.4$

$p(E)=0.45$

$p(J)=0.5$

$p(M\cap E)=0.2$

$p(M\cap E\cap J)=0.06$

We first need to determine,

$p(E\cap J)=p(E|J)*p(J)=0.4*0.5=0.02$

$p(M\cap J)=p(J|M)*p(M)=0.4*0.4=0.16$

Since we now have the required probabilities then,

$p(exactly\space one \space event)=0.4+0.45+0.5-2(0.2+0.02+0.16)+3(0.06)=0.59+0.18=0.77$

Therefore, the probability that Dorothy will join exactly one club is 0.77.