# Given that 14 mortgage foreclosures are audited by a large lending institution, find the probability of the following, below.Of all the mortgage foreclosures in the united stats, 61% are caused by...

Given that 14 mortgage foreclosures are audited by a large lending institution, find the probability of the following, below.

Of all the mortgage foreclosures in the united stats, 61% are caused by disability.People who are injured or ill cannot work, they then lose their jobs and thus their incomes. With no income, they cannot make their mortgage payment and the bank forecloses. Given that 14 mortgage foreclosures are audited by a large lending institution, find the probability of the following.

A. No more than 3 of the foreclosures are due to a disability

B. At least 2 of the foreclosures are due to a disability

C. exactly 9 of the foreclosures  are due to disability

D. At most 11 of the foreclosures are due to a disability

Asked on by mlunar74

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embizze | High School Teacher | (Level 1) Educator Emeritus

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We are given that `p=.61` so `q=.39` ,(p is the probability that a foreclosure is due to disability, q the probability that it is not caused by disability) and `n=14` (The number of samples taken).

(1) Find `P(x<=3)`

The actual probability is found by `P(x=0)+P(x=1)+P(x=2)+P(x=3)` where `P(x=a)=""_nC_ap^aq^(n-a)` ** (`_nC_a` is the number of combinations of n items taken a at a time -- this is also written `([n],[a])` )

So `P(x<=3)=""_(14)C_0(.61)^0(.39)^14+_(14)C_1(.61)^1(.39)^(13)+_(14)C_2(.61)^2(.39)^12+_(14)C_3(.61)^3(.39)^11` `~~.000002+.00004+.00042+.00262~~.0031`

The probability that at most 3 are due to disability is approximately .0031

(2) `P(x>1)` : It is far easier to find `1-P(x<=1)` , especially since we have calculated those.

So `P(x>1)=1-P(x<=1)=1-(.000002+.0004)~~.9996`

The probability that at least 2 are due to disability is approximately .9996

(3) `P(x=9)` : `_(14)C_9(.61)^9(.39)^5=.2112`

So the probability that exactly 9 are due to disability is approximately .2112

(4) `P(x<=11)` : Again it is easier to compute `1-P(x>11)` :

`1-[_(14)C_(12)(.61)^12(.39)^2+_(14)C_(13)(.61)^13(.39)^1+_(14)C_(14)(.61)^14(.39)^0]`

`~~1-[.0367+.0088+.0010]~~.9535`

So the probability of at most 11 being due to disability is approximately .9535

Sources:

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