binomial probability distribution

Statistics
Tutor: None Selected Time limit: 1 Day

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

Are you studying on the go? Check out our FREE app and post questions on the fly!
Download on the
App Store
...
Feb 27th, 2015
...
Feb 27th, 2015
Dec 5th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer