probability! its due by wednesday! please help!?

Question

i dont understand how to do this!? can someone please help me out.. its due in 2 daysss!!

10% of the population is left-handed. find the following probilities..
a) out of 4 people selected, at least 2 are left-handed
b) out of 6 people exactly 3 are left handed
c) if the % is actualy 11.3%, what is the probability hat if you pick 5 people, al will be left handed?

Answer 1

Let X be the number of people who are left handed. X has the binomial distribution.

In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.

this is found by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.
Or, to be more accurate, the binomial is the sum of n independent and identically distributed Bernoulli trials.

the mean of the binomial distribution is n * p
the variance of the binomial distribution is n * p * (1 - p)

for question a)
n = 4, p = 0.10
P(X ≥ 2)
= P(X = 2) + P(X = 3) + P(X = 4)
= 0.0486 + 0.0036 + 0.0001
= 0.0523

b)
n = 6, p = 0.10

P(X = 3) = 0.01458

c)

n = 5, p = 0.113

P(X = 5) = 1.842435e-05

Answer 2

a) The probability that exactly two are left-handed is (1/10)^2*(9/10)^2; that is, two are lefties with prob 1/10, and two are righties, each with prob 9/10. Similarly, exactly three are left-handed with prob (1/10)^3*(9/10), and all 4 are lefties with prob (1/10)^4. Therefore, at least two are lefties with prob 81/10^4 + 9/10^4 +1/10^4 = 91/10^4.

b) (1/10)^3*(9/10)^3 = 729/10^6.

You get the idea yet?