Question ID: #725
Two distinct numbers $a$ and $b$ are selected at random from $1, 2, 3, \dots, 50$. The probability that their product $ab$ is divisible by 3 is
- (1) $\frac{561}{1225}$
- (2) $\frac{664}{1225}$
- (3) $\frac{272}{1225}$
- (4) $\frac{8}{25}$
Solution:
Total number of ways to select 2 distinct numbers from 50 is:
$$ n(S) = {}^{50}C_2 = \frac{50 \times 49}{2} = 1225 $$
For the product $ab$ to be divisible by 3, at least one of the numbers $a$ or $b$ must be a multiple of 3.
It is easier to find the probability that the product is NOT divisible by 3 and subtract it from 1.
$$ P(\text{divisible by 3}) = 1 – P(\text{not divisible by 3}) $$
The product $ab$ is NOT divisible by 3 if both $a$ and $b$ are NOT divisible by 3.
Number of multiples of 3 in $\{1, 2, \dots, 50\}$:
$$ 3, 6, 9, \dots, 48 $$
This is an AP where $3n \le 50 \Rightarrow n = 16$.
So, there are 16 multiples of 3.
Number of non-multiples of 3:
$$ 50 – 16 = 34 $$
Number of ways to select 2 numbers from these 34 non-multiples:
$$ n(E’) = {}^{34}C_2 = \frac{34 \times 33}{2} = 17 \times 33 = 561 $$
Probability that product is NOT divisible by 3:
$$ P(E’) = \frac{561}{1225} $$
Required Probability:
$$ P(E) = 1 – \frac{561}{1225} $$
$$ P(E) = \frac{1225 – 561}{1225} = \frac{664}{1225} $$
Ans. (2)
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