SOLUTION: Please can someone help me? My homework is due soon and I am not sure what to do. These are my last two questions. 1) The probability that federal income tax returns will have

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Question 134510: Please can someone help me? My homework is due soon and I am not sure what to do. These are my last two questions.
1) The probability that federal income tax returns will have 0,1, or two errors is 0.53,0.15, and 0.3 respectively. If 10 randomly selected returns are audited, what is the probablity that six will have no errors, three will have one error, and one will have two errors?
A)0.184
B) 0.127
C) 0.020
D) 0.035

2) The probability that a person will have 0,1,or two dental checkups per year is 0.2,0.5, and 0.3 respectively. If seven people are picked at random, what is the probablity that two will have no checkup, four willl have one checkup and one will have two checkups in the next year?
A) 0.079
B) 0.013
C) 0.008
D) 0.043
Any and all help will be greatly appreciated. I am not sure at all how to approach these questions and to be honest, I really don't understand them.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1) The probability that federal income tax returns will have 0,1, or two errors is 0.53,0.15, and 0.3 respectively. If 10 randomly selected returns are audited, what is the probablity that six will have no errors, three will have one error, and one will have two errors?
P = [10!/(6!*3!*1!)]
= [840] is approximately 0.0184
--------------
A)0.184
B) 0.127
C) 0.020
D) 0.035
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2) The probability that a person will have 0,1,or two dental checkups per year is 0.2,0.5, and 0.3 respectively. If seven people are picked at random, what is the probablity that two will have no checkup, four willl have one checkup and one will have two checkups in the next year?
P = [7!/(2!*4!*1!]
= 105
= 0.07875
-----------------
A) 0.079
B) 0.013
C) 0.008
D) 0.043
=================
These are multinomial propability problems.
Use Google to find a nice explanation of how to work them.
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Cheers,
Stan H.