Question

A,B and C are events such that $P\left(A\right)=0.3,P\left(B\right)=0.4,P\left(C\right)=0.8,P(A\cap B)=0.08,P(A\cap C)=0.28$and $A\cup B\cup C)\ge 0.75$ show that $P(B\cap C)$ lies in the interval [0.23, 0.48]

Or

An integer is chosen random from 1 to 50, what is the probability that the integer chosen, is a multiple of 2 or 3 or 10?

Answer 1

We know that, the probability of occurrence of an event is always less than to 1 and it given that $P(A\cup B\cup C)\ge 0.75$

$\therefore 0.75P(A\cup B\cup C)\le 1$

$\Rightarrow 0.75\le P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cup B)-P(B\cup C)-0.28+0.09\le 1$

$\Rightarrow 0.75\le 0.3+0.4+0.8-0.08-P(B\cup C)-0.28+0.09\le 1$

$\Rightarrow 0.75\le 159-0.36-P(B\cap C)\le 1$

$\Rightarrow 0.75\le 1.23\le -P(B\cap C)\le 1$

$\Rightarrow 0.75-1.23\le -P(B\cap C)\le 1-1.23$

$\Rightarrow -0.48\le -P(B\cap C)\le -0.23$

$\Rightarrow 0.23\le P(B\cap C)\le 0.48$

Hene,$P(B\cap C)$ lies in the interval $[0.23,0.48].$

Or

Out of 50 integers an integer can be chosen in ${}^{50}{C}_{1}ways.$

$\therefore$Total number of elementary events,$n\left(S\right){=}^{50}{C}_1=50$

Consider the following events:

A=getting a multiple of 2

B=getting a multiple of 3

C=getting a multiple of 10

Now, $A\{2,4,6,8,...50\}$

$B\{3,6,9,...48\}$

$C\{10,20,30,40,50\}$

$A\cap B=\{6,12,18,...48\}$

$B\cap C=\left\{30\right\}$

$A\cap C=\{10,20,30,40,50\}$

and $A\cap B\cap C=\left\{30\right\}$

Then, $n\left(A\right)=25,n\left(B\right)=16,n\left(C\right)=5,n(A\cap B)=8$

$n(B\cap C)=1,n(A\cap C)=5andn(A\cap B\cap C)=1$

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{25}{50}$

$P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{16}{50}$

$P\left(C\right)=\frac{n\left(C\right)}{n\left(S\right)}=\frac{5}{50}$

$P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{8}{50}$

$P(B\cap C)=\frac{n(B\cap C)}{n\left(S\right)}=\frac{1}{50}$

$P(A\cap C)=\frac{n(A\cap C)}{n\left(S\right)}=\frac{5}{50}$

$P(A\cap B\cap C)=\frac{n(A\cap B\cap C)}{n\left(S\right)}=\frac{1}{50}$

$\therefore$ Required probability

$=P(A\cap B\cap C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)$

$=\frac{25}{50}+\frac{16}{50}+\frac{5}{50}-\frac{5}{50}+\frac{1}{50}=\frac{33}{50}$