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From a bag of 6 candies: 4 sweet, 2 sour Jack picks 2 candies simultaneously. What is the probability that first candy is sweet, second is sour?

My book says 4/6 x 2/5 = 4/15

The part I don't understand is why picking simultaneously and picking in a sequence are considered the same. There should be a difference.

Thanks for any replies.

Perhaps you should be asking - if you're picking them simultaneously, why is there a first and second? Sorry, that wasn't helpful.

Imagine if the first candy you pick is sour, and then the second candy you pick is sweet. Then what's the probability? It should be 2/6 x 4/5 = 4/15.

As you can see, the order doesn't actually matter. So whether they're picked simultaneously or picked in a sequence, the answer is the same.

The reason why they specified "simultaneously" is probably because they don't want you throwing the candy back into the bag after you pick it. (probability with replacement is different than probability without replacement) _________________

Hmm... So what about this question? Is it 1/9 or 2/9. I'm so confused.

There are 3 red, 2 white, and 5 blue marbles. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marbles?

Hmm... So what about this question? Is it 1/9 or 2/9. I'm so confused.

There are 3 red, 2 white, and 5 blue marbles. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marbles?

So what's the probability that the first marble we pick is white? There are 2 white marbles among 10 total marbles, so the probability is 2/10. After we pick out this white marble, what's the probability that the next marble we pick is blue? There are 5 blue marbles from 9 total marbles, so the probability is 5/9.

So, 2/10 x 5/9 = 1/9.

If we reverse the order of picking, and pick the blue marble first, then it becomes 5/10 x 2/9 = 1/9.

Hmm... So what about this question? Is it 1/9 or 2/9. I'm so confused.

There are 3 red, 2 white, and 5 blue marbles. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marbles?

So what's the probability that the first marble we pick is white? There are 2 white marbles among 10 total marbles, so the probability is 2/10. After we pick out this white marble, what's the probability that the next marble we pick is blue? There are 5 blue marbles from 9 total marbles, so the probability is 5/9.

So, 2/10 x 5/9 = 1/9.

If we reverse the order of picking, and pick the blue marble first, then it becomes 5/10 x 2/9 = 1/9.

Hmm... So what about this question? Is it 1/9 or 2/9. I'm so confused.

There are 3 red, 2 white, and 5 blue marbles. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marbles?

So what's the probability that the first marble we pick is white? There are 2 white marbles among 10 total marbles, so the probability is 2/10. After we pick out this white marble, what's the probability that the next marble we pick is blue? There are 5 blue marbles from 9 total marbles, so the probability is 5/9.

So, 2/10 x 5/9 = 1/9.

If we reverse the order of picking, and pick the blue marble first, then it becomes 5/10 x 2/9 = 1/9.

So same answer, either way.

When we add, the result is 2/9 though. Right?

Posted from my mobile device

Sorry if I've confused you. The answer is 1/9. I only did it both ways to show that order of sequence does not matter. There's no reason to add anything. _________________

But there are C(10,2)=45 ways to choose 2 balls out of 10. There are 2x5=10 desired out comes. 10/45 = 2/9 I'm so confused now. If the answer to this question is 1/9, I have some really serious problems.

But there are C(10,2)=45 ways to choose 2 balls out of 10. There are 2x5=10 desired out comes. 10/45 = 2/9 I'm so confused now. If the answer to this question is 1/9, I have some really serious problems.

Ugh, I'm terribly sorry. You are correct, the answer should be 2/9.

We should add the 1/9 and 1/9 I got from my previous answers because there are two ways to get 1 white and 1 blue... choosing the white first then blue, or choosing the blue first then white.

I feel I may have made this whole thing more confusing for you. My apologies. _________________

No problem farful, thank you for your replies. Since I'm not a native speaker, now all simultaneous picking, picking in order, etc... are much more clear. Cheers.

From a bag of 6 candies: 4 sweet, 2 sour Jack picks 2 candies simultaneously. What is the probability that first candy is sweet, second is sour?

My book says 4/6 x 2/5 = 4/15

The part I don't understand is why picking simultaneously and picking in a sequence are considered the same. There should be a difference.

Thanks for any replies.

Let me reanswer this.

The reason why they said 'simultaneously' is because they wanted you to pick one candy, then pick a second candy without putting the first candy back into the bag.

So they really want the probability of picking a sweet candy first, followed by the probability of picking a sour candy next (from the remaining 5).

If the question was: From the bag, pick two candies, what is the probability one is sweet and one is sour? Then, the answer would be 8/15. You can get this answer either 1) via (4x2)/(2choose6) or 2) figuring out the probability of picking sweet then sour and adding that to the probability of picking sour then sweet, since there are two ways to get one sweet and one sour.

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