When probability's a chaos

Probability Level pending

You are given two sets of numbers $M$ and $N$ . $M$ consists of numbers from $\{0, 2, 4, 6, 8\}$ and $N$ consists of numbers from $\{1, 3, 5, 7, 9\}$ . Two numbers are chosen at a random from $M$ and $N$ one each.

Let the possibility of getting a multiple of $3$ when the numbers chosen at random are added be $\dfrac{a_1}{b_1}$ , where $a_1$ and $b_1$ are coprime integers.
Let the possibility of getting a natural number when numbers chosen are subtracted, ie $M-N$ , be $\dfrac{a_2}{b_2}$ , where $a_2$ and $b_2$ are coprime integers.
Let the possibility of getting an integer when $M$ is divided by $N$ be $\dfrac{a_3}{b_3}$ , where $a_3$ and $b_3$ are coprime integers.

Find the value of $b_1 - b_2 - b_3 - a_1 - a_2 - a_3$

The answer is 2.

1 solution

Viki Zeta
Oct 26, 2016

When $M$ and $N$ are chose at the same time, below are the following possibilities.

$(0,1), (0,3), (0,5), (0,7), (0,9)\\ (2,1), (2, 3), (2, 5), (2, 7), (2, 9)\\ (4,1), (4,3), (4,5), (4,7), (4,9)\\ (6,1), (6,3), (6,5), (6,7), (6,9)\\ (8,1), (8,3), (8,5), (8,7), (8,9)\\$

So only 9 sets, $\{ (0, 3), (0, 9), (2, 1), (2, 7), (4, 5), (6, 3), (6, 9), (8, 1), (8, 7) \}$ , add up to give a multiple of 3.

$\therefore$ Probability of getting sum a multiple of 3 = $\dfrac{9}{25} = \dfrac{a_1}{b_1}$

Only these numbers $\{ (2, 1), (4, 1), (4, 3), (6, 1), (6, 3), (6, 5), (8, 1), (8, 3), (8, 5), (8, 7) \}$ result in a natural number when subtracted.

$\therefore$ Probability of getting difference a natural number = $\dfrac{10}{25} = \dfrac{2}{5} = \dfrac{a_2}{b_2}$

These numbers $\{(0, 1), (0, 3), (0, 5), (0, 7), (0, 9), (2, 1), (4, 1), (6, 1), (6, 3), (8, 1)\}$ when divided result in a integer.

$\therefore$ Probability of getting a integer when divided = $\dfrac{10}{25} = \dfrac{2}{5} = \dfrac{a_3}{b_3}$

$\boxed{\therefore b_1 - a_2 - a_3 - b_2 - b_3 = 25 - 5 - 5 - 2 - 2 - 9 = 2}$

When probability's a chaos

The answer is 2.

1 solution

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