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Expert: - 4/24/2011


Question
QUESTION: Approximately 76% of families in Toronto consist of married couples. There are 22 students in a grade3 class. The school administration records whether they live with both biological parents who are either married or living common-law.If the child is from a single-parent home, they record the parent with which the child lives.
---What is the probability that exactly half the students have families where the parents are defined as a married couple? Express your answer to the nearest tenth of a percent.

ANSWER: Since there are 22 students and p = 0.76, 1-p = 0.24.

To have exactly half living with a couple would mean that we had 11 with a couple
and 11 with only one parent.

This follows the binomial distribution and the value would be C(22,11)(0.76^11)(0.24^11).
Note that C(a,b) = a!(b!(a-b)!).

Using Excel, I get
]1.52168E-07   0.76^11
0.048859556   0.24^11
705432      22!/(11!11!) { (22*21*20*19*18*17*16*15*14*13*12)/(11*10*9*8*7*6*5*4*3*2*1) }
0.005244793   result of multiplying the above three values

That would be 0.5%.


---------- FOLLOW-UP ----------

QUESTION: What is the probability that atleast one student doesnot have parents that are defined as a married couple?

Answer
Since the probability of living with a married couple is 0.76
and there are 22 students, it would be (0.76)^22, which is 0.002387256.

To get some number between 1 and 22 students with married parents would be
0   2.31551E-14
1   1.61314E-12
2   5.36369E-11
3   1.13234E-09
4   1.70322E-08
5   1.94167E-07
6   1.74211E-06
7   1.26096E-05
8   7.48693E-05
9   0.000368801
10   0.001518229
11   0.005244793
12   0.015224468
13   0.037085242
14   0.075494957
15   0.127502594
16   0.176644219
17   0.197425892
18   0.173661664
19   0.115774443
20   0.05499286
21   0.016585148
22   0.002387256

The sum of all these probabilities is 1, as it should be.

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