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I've been trying to figure out how they got a (15) from (6/2).
Q: suppose that 55% of all babies born in a particular hospital are girls. If 6  babies born in the hospital are randomly selected, what is the probability that fewer than 3 of them are girls?

A:step 1 : (6/0)(0.55)^0(1-0.55)^6+(6/1)(0.55)^1(1-0.55)^5=(6/2)(0.55)^2(1-0.55)^4
(step 2: 1)(0.55)^0(1-0.55)^6+ (6)(0.55)^1(1-0.55)^5+(15)(0.55)^2(1-0.55)^4
=0.2553
I don't get this at all

The solution looks OK to me. Lets make sure we agree on the notation and conventions to answer your question about "(15) from (6/2)".

For the binomial distribution, the notation for choosing the number of combinations of m objects out of n is written

C(n,m) = ( n )
( m )

where the parentheses represent one big parenthesis bracketing the letter m placed directly below the letter n (not divided by!). I believe this is what you intended with your expression (6/2), which, given the shortcomings of the text editor we are using, is reasonable.

C(n,m) stands for the expression

C(n,m) = n! / { m!･(n-m)! }

where ! is the factorial notation meaning n! = n･(n-1)･(n-2)...(2)･(1). For your case

C(6,2) = ( 6･5･4･3･2･1) / { 4･3･2･1 )･( 2･1 ) } = ( 6･5 ) / 2 = 30/2 = 15.

The solution thus consists of adding up the probabilities of getting 0, 1 and 2 girls out of 6 using the binomial distribution formula.

Hope this helps.

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#### randy patton

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college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

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