Question

In 2014 the Department of Social Services reported that 32% of current marriages in Australia were expected to end in divorce.
Find the probability that more than 8 marriages out of a random sample of 20 marriages which were current in 2014 would end in divorce.

Answer 1

Answer on Question #55689 – Math – Statistics and Probability

Question

In 2014 the Department of Social Services reported that 32% of current marriages in Australia were expected to end in divorce.

Find the probability that more than 8 marriages out of a random sample of 20 marriages which were current in 2014 would end in divorce.

Solution

We need to find the following probability: $p(>8 \text{ divorces})$.

Obviously, more than 8 divorces means either 9 divorces or 10 divorces or 11 divorces and so on and so forth:

$$\begin{array}{l} p(>8 \text{ divorces}) = p(9 \text{ divorces}) + p(10 \text{ divorces}) + p(11 \text{ divorces}) + \cdots + \\ + p(20 \text{ divorces}). \end{array}$$

Now, let us find the probability of exactly $k$ divorces.

A divorce happens with a probability $p = 0.32$.

If we have $k$ divorces, then we have $(20 - k)$ happy couples not divorced. Assume that couples 1 through $k$ have divorced, while others have not. It means that $1^{\text{st}}$ marriage ended with a divorce (with probability $p$), and the $2^{\text{nd}}$ one ($p$ again), ..., while $(k+1)^{\text{th}}$ did not end with a divorce (with probability $1 - p$), ... These events occurred simultaneously, so we should multiply their probabilities: $p^k (1 - p)^{n - k}$.

But it could've happened with the other combination of $k$ marriages chosen from 20, while there are $\binom{20}{k} = \frac{20!}{k!(20 - k)!}$ ways to choose $k$ couples from 20. Thus, we have

$$p(k \text{ divorces}) = \binom{20}{k} p^k (1 - p)^{n - k}.$$

Thus,

$$p(>8 \text{ divorces}) = \sum_{k=9}^{20} \binom{20}{k} p^k (1 - p)^{n - k}.$$

Substituting $p = 0.32$ we have

$$p(>8 \text{ divorces}) = \sum_{k=9}^{20} \binom{20}{k} 0.32^k 0.68^{20 - k} \approx 0.1568$$

(it was computed by means of Wolfram|Alpha Widgets: Binomial Distribution Calculator).

In Wolfram Mathematica we can also use

```

functbin[n_]:=BinomialDistribution[n,0.32];

Probability[9 ≤ x ≤ 20, x ≈ functbin[20]]

```

In Excel 2013 it can be calculated by means of the following expression:

$$= \text{BINOM.DIST.RANGE}(20;0,32;9;20)$$

In Excel 2010 and Excel 2013 it can be calculated by means of the following expression:

$$= \text{BINOM.DIST}(20;20;0.32;\text{TRUE}) - \text{BINOM.DIST}(8;20;0,32;\text{TRUE})$$

In Excel 2000, Excel XP, Excel 2003, Excel 2007, Excel 2010, Excel 2013 it can be calculated by means of the following expression:

=BINOMDIST(20;20;0.32;TRUE)-BINOMDIST(8;20;0,32;TRUE)

The meaning of the functions are the following:

BINOM.DIST.RANGE(n; p; x; y)=the probability there are between $x$ and $y$ successes (inclusive) in $n$ trials where the probability of success on any trial is $p$.

BINOMDIST(x; n; p; TRUE)=cumulative probability distribution F(x) value at x for the binomial distribution B(n, p), i.e. the probability that there are at most x successes in n trials where the probability of success on any trial is p. Here x is a non-negative integer, n is a positive integer, 0<p<1. A value of TRUE returns the cumulative distribution function.

BINOM.DIST is equivalent to BINOMDIST:

BINOM.DIST(number_success; number_trial; p; TRUE).

In R language it can be calculated by means of the following expression:

pbinom(20,20,0.32)-pbinom(8,20,0.32)

Answer: 0.1568.

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