binomial probability distribution

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Feb 27th, 2015

OK, we have a fair quarter, so p = 0.5 for heads or tails. And we flip the coin three times, so n=3. Each question gives us k (the number of times a given outcome occurs) so we can calculate the probability using the formula:

n!/(k!(n-k)!) * p^k * (1-p)^(n-k)

So for a) we have k = 3, so: 

3!/(3!(0)!) * 0.5^3 * 0.5^0

= 1 * 0.125 * 1

= 0.125 (or 1-in-8 which makes sense, right? 0.5^3 since we need 3 probability-0.5 events to all occur).

Similarly for b) we just do the same with k=2 instead, and we get:

3! / (2!(1)!) * 0.5^2 * 0.5^1

= 6 / 2 * 0.25 * 0.5 

= 0.375 (or 3/8).

For part c), since we know the probability of getting 2 or 3 heads, and they're independent, the probability of getting either (i.e. the probability of getting 2+ heads) is just their sum - 0.125+0.375 = 0.5.

d) This is exactly the same as part a), since the probability of tails is identical to the probability of heads (0.5).


Hope this helped!

Feb 27th, 2015

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