
Name: Amna
Who is asking: Parent
Level: Secondary
Question:
I had a few binomial probability questions which I can not use from the tables as instructed:
 If 60 % of television viewers are watching a certain program, what is the probability tha tmore than half of those selected in a random sample of five will be watching the specified program?
 If it is known according to Mendel's Law, that we can expect in teh long run to have 3 white, 1 brown rabbits in every 4 rabbits of a certain type, what is the probability that 2 in a litter of 3 will be white?
 On the average, 2% of the items sold in a department store are returned for refunds. what is the probability that of its next five items sold, at most two will be returned for refunds?
Thanks!
Hi there,
For the following problems, set them up as follows:
 If 60 % of television viewers are watching a certain program, what is the
probability tha tmore than half of those selected in a random sample of five
will be watching the specified program?
Define success as the viewers are watching the show
Then probability of success, p = 0.6 and q = 0.4
The number of trials is 5 so n = 5
Need to find the probability that more than half of the sample is watching the show. Since five is odd, more than half would be 3 or 4 or 5.
So find P(3 or 4 or 5) = P(3) + P(4) + (5)
 If it is known according to Mendel's Law, that we can expect in teh long run
to have 3 white, 1 brown rabbits in every 4 rabbits of a certain type, what is
the probability that 2 in a litter of 3 will be white?
Success = being white
p = 0.75, q = 0.25
n = 3
Find P(2)
 On the average, 2% of the items sold in a department store are returned for
refunds. what is the probability that of its next five items sold, at most two
will be returned for refunds?
Success = being returned
p = 0.02, q = 0.98
n = 5
Find P(0 or 1 or 2) = P(0) + P(1) + P(2)
Once you have them set up like this, use the definition of a binomial probability to calculate the individidual probabilities and then add them. Remember the formula for a binomial probability:
P(k) = nCk(p^{k})(q^{nk})
Hope this helps,
Leeanne
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