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# Probability Questions of Patients Bills

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1) Records show that 30% of all patients admitted to a medical clinic fail to pay their bills and that eventually the bills are forgiven. Suppose n = 4 new patients represent a random selection from the large set of prospective patients served by the clinic. Find these probabilities:
a) All the patients' bills will eventually have to be forgiven
b) One will have to be forgiven
c) Nine will have to be forgiven

2) During the 1992 football season, the Los Angeles Rams had a bizarre streak of coin-toss losses. In fact, they lost the call 11 weeks in a row.
a) The Rams' computer system manager said that the odds against losing 11 straight tosses are 2047 to 1. Is he correct?
b) After these results were published, the rams lost the call for the next two games, for a total of 13 straight losses. What is the probability of this happening if, in fact, the coin was fair?

3) Tay-Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carriers of the disease, the probability that their offspring will develop the disease is approximately .25. Suppose a husband and wife are both carriers and the wife is pregnant on three different occasions. If the occurrence of Tay-Sachs in any one offspring is independent of the occurrence in any other, what are the probabilities of these events?
a) All three children will develop Tay-Sachs disease
b) Only one child will develop Tay-Sachs disease
c) The third child will develop Tay-Sachs disease; given the first two did not

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1) Records show that 30% of all patients admitted to a medical clinic fail to pay their bills and that eventually the bills are forgiven. Suppose n = 4 new patients represent a random selection from the large set of prospective patients served by the clinic. Find these probabilities:
a) All the patients' bills will eventually have to be forgiven.
b) One will have to be forgiven
c) Nine will have to be forgiven.

This is Binomial probability distribution
a) All the patients' bills will eventually have to be forgiven

P(r)= ncr p r * q n-r

n= 4
p= 0.3 or 30%
q=1-p= 0.7 or 70%
P(r)= ncr p r * q n-r =
When r=4 , P(4)= 0.0081 or 0.8100% =1*(0.3^4)*(0.7^0)
Probability = 0.0081 or 0.8100%
(Note ^ indicates raised to the power of)

b) One will have to be forgiven

Here r=1
P(r)= ncr p r * q n-r

n= 4
p= 0.3 or 30%
q=1-p= 0.7 or 70%

P(r)= ncr p r * q n-r =
when ...

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