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How to work out binomial distribution?
I really do not undersand normal distribution and I need it for my Stats. I've got this question and if anyone could talk it through for me that would be great thanks!
Megan throws darts at a target. She can hit the target 7 times in 9 throws. She throws 3 darts.
Assuming a binomial distribution, work out the probability that she hits the target exactly twice. You may use:
(p+q)^3=p^3+3p^2q+3pq^2+q^3
Thankyou!
Hi,
She can hit the target 3,2,1, or 0 times. There are 2³ or 8 possible outcomes, which divide up into the binomial expansion coefficients for each result.
3 hits can occur only one way.
2 hits can occur 3 ways, YYN, YNY, or NYY.
1 hit can occur 3 ways, YNN, NYN, or NNY.
0 hits occur one way, NNN
When the odds of her success on a throw is 7/9 then the odds of missing is 2/9. The results could be:
3nCr3(7/9)^3(2/9)^0 = .011 = 1.1% probability of 3 hits
3nCr2(7/9)^2(2/9)^1 = .115 = 11.5% probability of 2 hits
3nCr1(7/9)^1(2/9)^2 = .403 = 40.3% probability of 1 hit
3nCr0(7/9)^0(2/9)^3 = .471 = 47.1% probability of 0 hits
3nCr3 = 1
3nCr2 = 3
3nCr1 = 3
3nCr0 = 1
These are the binomial coefficients above. They are multiplied times the 7/9 likelihood raised to the number of hits times the 2/9 likelihood of missing the target raised to the number of misses.
3nCr2(7/9)^2(2/9)^1 = .115 = 11.5% probability of 2 hits
There is an 11.5% probability of 2 exactly hits.
I hope that helps!!
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