Question 1124808: Assume that 64% of people are left-handed. If we select 5 people at random, find the probability of each outcome described below, rounded to four decimal places:
a. There are some lefties (≥ 1) among the 5 people.
b. There are exactly 3 lefties in the group.
c. There are at least 4 lefties in the group.
d. There are no more than 2 lefties in the group.
e. How many lefties do you expect?
f. With what standard deviation?
Please help I am very confused! Especially with the problem E and A.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Some lefties (greater than or equal to 1) has probability 1- no lefties, or 1-.64^5=0.8926
Exactly 3 is 5C3*0.64^3*0.36^2, the 5C3 being number of ways (10) to choose 3 from 5.
probability is 0.3397
Exactly 4 is 5*0.64^4*0.36=0.3020
Exactly 5 is 0.64^5=0.1074
Exactly 2 is 10*0.64^2*0.36^3=0.1911
Exactly 1 is 5*.64*.36^4=0.0537
0 is .36^5=0.0060
Using these, at least 4 includes 4 or 5. and that has probability 0.4094
no more than 2 means 0,1,2, and that probability is 0.2508
Expected value is n*p or 5*0.64=3.2
sd is sqrt (variance), and variance is np(1-p)=3.2*0.36=1.152. sqrt (1.152)=1.07, the sd.
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