An integer is chosen at random from the integers [1,2,3, , 50].

Question

An integer is chosen at random from the integers $$1,2,3, \ldots, 50$$. The probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is

Accepted Answer

Given set $$=\{1,2,3, \ldots \ldots . .50\}$$

$$\mathrm{P}(\mathrm{A})=$$ Probability that number is multiple of 4

$$\mathrm{P(B)}=$$ Probability that number is multiple of 6

$$\mathrm{P}(\mathrm{C})=$$ Probability that number is multiple of 7

Now,

$$\mathrm{P}(\mathrm{A})=\frac{12}{50}, \mathrm{P}(\mathrm{B})=\frac{8}{50}, \mathrm{P}(\mathrm{C})=\frac{7}{50}$$

again

$$\begin{aligned}
& \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{50}, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{1}{50}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{50} \
& \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0
\end{aligned}$$

Thus

$$\begin{aligned}
P(A & \cup B \cup C)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}+0 \
& =\frac{21}{50}
\end{aligned}$$

Solutions · 1