Question
Suppose an integer from $1$ through $1000$ is chosen at random$,$ find the probability that the integer is a multiple of $2$ or a multiple of $9.$
✓
Answer
We have integers $1, 2, 3, …., 1000$
We have integers $1, 2, 3, ….., 1000$
$n(S) = 1000$
Number of integers which are multiple of $2 = 500$ Let the number of integers which are multiple of $9$ be $n.$
$n^{th} term = 999$
$\Rightarrow 9 + (n -1)9 = 999$
$\Rightarrow 9 + 9n - 9 = 999$
$\Rightarrow n = 111$
From $1$ to $1000,$ the number of multiples of $9$ is $111.$
The multiple of $2$ and $9$ both are $18, 36, …., 990.$
Let m be the number of terms in above series.
$m^{th}\ term = 990$
$\Rightarrow 18 + (m - 1)18 = 990$
$\Rightarrow 18 + 18m - 18 = 990$
$\Rightarrow m = 55$
Number of multiples of $2\ or\ 9 = 500 + 111 - 55 = 556 = n(E)$
$\therefore\ \text{Required probability}=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}$
$=\frac{556}{1000}=0.556$
We have integers $1, 2, 3, ….., 1000$
$n(S) = 1000$
Number of integers which are multiple of $2 = 500$ Let the number of integers which are multiple of $9$ be $n.$
$n^{th} term = 999$
$\Rightarrow 9 + (n -1)9 = 999$
$\Rightarrow 9 + 9n - 9 = 999$
$\Rightarrow n = 111$
From $1$ to $1000,$ the number of multiples of $9$ is $111.$
The multiple of $2$ and $9$ both are $18, 36, …., 990.$
Let m be the number of terms in above series.
$m^{th}\ term = 990$
$\Rightarrow 18 + (m - 1)18 = 990$
$\Rightarrow 18 + 18m - 18 = 990$
$\Rightarrow m = 55$
Number of multiples of $2\ or\ 9 = 500 + 111 - 55 = 556 = n(E)$
$\therefore\ \text{Required probability}=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}$
$=\frac{556}{1000}=0.556$
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