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# S2(1)

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1. Hi, question on the binomial distribution again!

I would like to know why if a probability asks for the probability of a success that we count the failures too?

For example: A fair die is rolled 7 times. Find the probability of getting

(b) only 3 fives

I know the answer will be

but I want to know why we multiply the last bit ? It only asks for the probabilitiy of 3 fives so I don't don't understand why count the probability of not getting 3 fives 4 times
2. ellooo
3. This is just how the binomial distribution handles the 'counting' involved. You want 3 of the rolls to be fives and 4 NOT to be fives. The last bit is because you don't want the other 4 rolls to be fives. The first bit comes from the counting (how many possible such combinations are there?).
4. (Original post by Naruke)
Hi, question on the binomial distribution again!

I would like to know why if a probability asks for the probability of a success that we count the failures too?

For example: A fair die is rolled 7 times. Find the probability of getting

(b) only 3 fives

I know the answer will be

but I want to know why we multiply the last bit ? It only asks for the probabilitiy of 3 fives so I don't don't understand why count the probability of not getting 3 fives 4 times
5. (Original post by Zacken)
Noooo

You told me about the n chose r meaning

Now I'm confused about the second probability. why does the probability of failure get multiplied too?
6. (Original post by Naruke)
Noooo

You told me about the n chose r meaning

Now I'm confused about the second probability
Okay.
7. (Original post by Zacken)
Okay.
Can u explain plzz

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