Math, asked by jbnklita2ty, 10 months ago

two numbers are selected at random from a set of first 90 natural numbers. find the probabilty that the product of randomly selected numbers is divisible by 3

Answers

Answered by parvathykm2002

Answer:

1/3

Step-by-step explanation:3*30=90

so 30 would be the probability

now 30/90 gives 1/3......

Answered by VedaantArya

Answer:

5/9 = 0.555... (if repetition is allowed)

Step-by-step explanation:

0. I'll be proceeding by selecting the two numbers in an ordered way.

1. If the first number is divisible by 3, well, it doesn't matter what the second number is. The probability of this happening is: 1/3. (30 numbers in the first 90 which are divisible by 3)

2. Now, if, say, the first wasn't (whose chance is 2/3), the chances of the second number being so is also 1/3, since repetition is... enabled (let). The chance of this whole event happening is hence 2/3 * 1/3 = 2/9.

3. The total probability is 1/3 + 2/9 = 5/9.

The answer shall be different, and easy to solve but difficult to look at, if repetition wasn't allowed. In this case though, I'd directly assume it was.

guddu61293: But multiplication is commutative. So selecting one number which is divisible by 3 is enough. Do we need the second case? Please shed some light on the second case. I am a confused.

VedaantArya: First, I'll prove that the 2nd case is absolutely necessary, and that the above solution is correct: Let the product be AB. We are now going to select the numbers A and B. The event that AB is divisible by 3, has probability = 1 - AB is not divisible by 3. And AB is not divisible by 3 when both of A and B aren't. The probably of BOTH of them not being divisible by 3 is (2/3) * (2/3) = 4/9. Hence the probability that AB IS divisible by 3, is 1 - 4/9 = 5/9.

VedaantArya: Next, about your particular statement (Line 2: So selecting...).
Yes it is, but there are more cases: when the one number, and be careful about this, that we are talking about isn't divisible by 3, the other number, which we aren't talking about, might be. This is the extra probability which we aren't taking care of.

VedaantArya: To elaborate further: When we say we're selecting one number which is divisible by 3, we have decided that we are speaking of particularly A or particularly B in the product AB. That's what "one number" says. So, the cases which correspond to the other number ever* being divisible by 3, and hence rendering the product AB satisfactory - they need to be counted. This is not equivalent to re-counting them. (*: not exactly ever, since the ones in which both A, B are / by 3, are already counted)

VedaantArya: If this still doesn't clear it, kindly try to re-state your query in a different way.

Previous Question

Next Question