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My approach is that I have total of $\frac{100}{3} = 33$ numbers that are divisible by 3.

Now, when I pick 1, and I encounter any of this 33 numbers like (1,3), (1,6),... their product is divisible by 3.

Then again when I pick 2, I encounter this 33 numbers like (2,3), (2,6),... their product is divisible by 3.

So, out of this 67 numbers (not divisible by 3), I have 67*33 cases where it becomes divisible by 3.

Again when I choose the 33 numbers like (3,6,9..) and associate each of them them with rest 99 numbers, the product is divisible by 3

Ex: (3,1), (3,2), (3,4).. (3,100)

So, I have 67*33 + 33*99 cases = 5478

Therefore probability becomes $\frac{5478}{^{100}P_2} = 0.55$

Is my logic correct? Any help is appreciated. Thanks in advance.

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  • $\begingroup$ I think you're right because also you can take all the pairs which product isn't divisible by 3 ($67\cdot 67$) and then just calculate $1-\frac{67\cdot 67}{100\cdot 100}$ $\endgroup$ Oct 21, 2016 at 6:39
  • $\begingroup$ Why $^{100}P_2$? It should be $^{100}C_2$, shouldn't it? $\endgroup$ Oct 21, 2016 at 6:39
  • $\begingroup$ @астон вілла олоф мэллбэргo.. like I'm taking all probable cases.. (1,3) is not (3,1).. Like that.. So, it's arrangement.. Not selection.. Am I wrong? $\endgroup$
    – lu5er
    Oct 21, 2016 at 6:41
  • $\begingroup$ Oh, I understood. Yes the answer is fine $\endgroup$ Oct 21, 2016 at 6:42
  • $\begingroup$ @ астон вілла олоф мэллбэрг.. Thanks. :) $\endgroup$
    – lu5er
    Oct 21, 2016 at 6:43

3 Answers 3

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Yes, you are right. Here's another way of thinking about it.

Let the numbers on the two cards be $x$ and $y$. Since $3$ is prime, $3$ divides $xy$ if and only if $3$ divides $x$ or $3$ divides $y$.

$$ \mathbb{P}(3 \mid x \text{ or } 3 \mid y) = 1 - \mathbb{P}(3 \nmid x)\mathbb{P}((3 \nmid y) \mid (3 \nmid x)) = 1 - \frac{67}{100}\frac{66}{99} = \frac{83}{150} $$

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  • $\begingroup$ Woow! Good one.. :) Easy and simple.. arberavdullahu also said like this in the comment, but he is wrong as he's taking replacements. Anyway thanks.. Nice explanation :) $\endgroup$
    – lu5er
    Oct 21, 2016 at 6:48
  • $\begingroup$ The first equality is not true, since $\mathbb{P}\left(3\nmid x\right)=\mathbb{P}\left(3\nmid y\right)=\frac{67}{100}$. You should write: $\mathbb{P}\left(3\mid x\text{ or }3\mid y\right)=1-\mathbb{P}\left(3\nmid x\text{ and }3\nmid y\right)=1-\mathbb{P}\left(3\nmid x\right)\mathbb{P}\left(\left[3\nmid y\right]\mid\left[3\nmid x\right]\right)=1-\frac{67}{100}\frac{66}{99}$ $\endgroup$
    – drhab
    Oct 21, 2016 at 8:16
  • $\begingroup$ If this is modified correctly then my downvote will turn into an upvote. $\endgroup$
    – drhab
    Oct 21, 2016 at 9:29
  • $\begingroup$ Sorry about that, thanks for the correction. $\endgroup$ Oct 22, 2016 at 0:21
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I approach questions like this from the opposite side, when is my card not divisible by three? Well that's when both my cards aren't divisible by 3. If I take a card, replace it and take another then there is a .67 chance on both, if I take a card, don't replace it and then take another, then there is a .67 chance and then a $2/3$ chance (66 not divisible, 99 total). Either way, we multiply these chances to find the chance my card isn't divisible by 3, and subtract this number from 1 to find the compliment (i.e. total is divisible by 3).

1-(.67.67)=.5511 (with replacement)

1-($\frac{67}{100} \frac{66}{99}$)=.55333...

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Because $3$ is a prime number, this question becomes quite easy. The product is divisible by $3$ if and only if at least one of the cards' values is divisible by $3$.

The chance that at least one of the two cards' values is divisible by $3$ is $$1 - P(\text{both not divisible by }3)$$

Calculating $P(\text{both not divisible by } 3)$:

In the $100$ cards, $\lfloor \frac {100} 3 \rfloor$ have values divisible by $3$. This means the first card has a $1 - \lfloor \frac {100} 3 \rfloor/100$ chance of not being divisible by $3$. This is $67/100$.

Given the first card is not divisible by $3$, there are still $\lfloor \frac {100} 3 \rfloor$ cards divisible by $3$, but one less card to choose from. This means the second card's chance to not be divisible by $3$ is $1 - \lfloor \frac {100} 3 \rfloor/99$ which is $2/3$.

Hence $$\begin{align} 1- P(\text{both not divisible by }3) &= 1 - \frac {67} {100} \cdot \frac 2 3 \\ &= 1 - \frac {134} {300} \\ &= \frac {166}{300} \\ &= \frac {83} {150}\end{align}$$ which is the required answer.

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