probability????.....not a fan

[Mel Mitch]

I had this problem as an assignment three months back....but i sumbitted without completing this question....:S.......now my test is in september and it should be loaded on probability :S....Help me study pleaseeeeeeeeeee

Problem
Out of a large population it may be assumed that 5% of unmarried men of forty years of age will marry within five years. Calculate the probability that out of a sample of 8 unmarried men who are 40 years old, selected at random
i) none will get married within 5 years
ii)just 3 will get married within 5 years
iii)atleast 4 will get married within 5 year

did i mention that i'm not a probability fan....

 

[tkhunny]

Expand \(\displaystyle (a+b)^{8}\)
Substitute a = 0.05 into each term separately.
Substitute b = 1-a = 0.95 into each term separately.
Resist the temptation to add the terms together.

The first term, \(\displaystyle a^{8}\), represents the probability that all 8 will marry.
The second term, \(\displaystyle 8a^{7}b\), represents the probability that all 7 will marry.
and so on.

This is a Binomial Expansion. You should be familiar with it from algebra. You can be required on an exam to calculate individual terms at the drop of a hat.

 

[soroban]

Hello, Mel Mitch!

You're expected to be familiar with Binomial Probabilities.
. . I'll try to explain in baby steps . . .

In a large population, 5% of single men of age 40 will marry within five years.

Calculate the probability that out of a sample of 8 single men of age 40 selected at random:

a) none will get married within 5 years

\(\displaystyle \text{We have: }\:\begin{array}{ccccc}P(M) &=& P(\text{married}) &=& 0.05\\ P(S) &=& P(\text{single}) &=& 0.95 \end{array}\)


\(\displaystyle \text{"None of the eight get married" means "all eight are single".}\)

. . \(\displaystyle P(\text{8S}) \:=\0.95)^8 \:=\:0.663420431\)

b) exactly 3 will get married within 5 years

\(\displaystyle P(\text{3M, 5S}) \;=\;{8\choose3}(0.05)^3(0.95)^5 \;=\; 0.005416467\)

c) at least 4 will get married within 5 years

\(\displaystyle \text{The opposite of "at least 4M" is "less than 4M".}\)
. . \(\displaystyle \text{That is: }\:0M,\:1M,\:2M,\text{ or }3M\)


\(\displaystyle \begin{array}{ccccccccc}\text{ 0M, 8S} &=& \text{from (a)} &=& 0.663 420 431 \\ \\[-3mm] \text{1M, 7S} &=& {8\choose1}(0.05)(0.95)^7 &=& 0.279 334 918 \\ \\[-3mm] \text{2M, 6S} &=& {8\choose2}(0.05)^2(0.95)^6 &=& 0.051 456 432 \\ \\[-3mm] \text{3M, 5S} &=& \text{from (b)} &=& 0.005 416 467 \\ \hline \\[-3mm] & & \text{Total: }& & 0.999 628 248 \end{array}\)


\(\displaystyle \text{Therefore: }\(\text{at least 4M}) \;=\;1 - 0.999 628 248 \;=\;0.000 371 752\)

 

[Mel Mitch]

wow....u make it prepare really easy....and that's exactly the steps that i was looking for......very clear.... for now i 'm liking probability
Thanks soroban