Converting from Individual probability to Household probability

May 2010
2
0
I think this is basic, but I just can't figure it out. Any help would be appreciated...

There is a 65% chance that each individual likes Product A.

For a household of 2 individuals, what is the chance that someone in the household likes Product A?

For a household of 3?

I would appreciate the formula in addition to the answer if possible. Thank you very much for your help.
 

mr fantastic

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Dec 2007
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I think this is basic, but I just can't figure it out. Any help would be appreciated...

There is a 65% chance that each individual likes Product A.

For a household of 2 individuals, what is the chance that someone in the household likes Product A?

For a household of 3?

I would appreciate the formula in addition to the answer if possible. Thank you very much for your help.
Let X be the random variable 'number of individuals in household who like product A'.

X ~ Binomial(n = total number of individuals in household, p = 0.65).

Calculate 1 - Pr(X = 0).
 
May 2010
2
0
Thank you for pointing me in the direction of binomial distributions!

I think I figured out that in order for me to get the total probability for the household, I must use that formula for each member of the household, and add up the results. For instance, for a 3 person household my excel formula looks like:
=BINOMDIST(1,3,.65,FALSE) +
BINOMDIST(2,3,.65,FALSE) +
BINOMDIST(3,3,.65,FALSE)

(The format for the BINOMDIST formula in Excel is
BINOMDIST(number_s, trials, probability_s, cumulative)
Where number_s = the number of successes in trials,
trials = the number of independent trials,
Probability_s = probability of success,
Cumulative - When FALSE returns probability that there are number_s successes.)

This method returns these results for the given number of individuals in the household:
1 = .65
2 = .87
3 = .95
4 = .98
5 = .99

This "looks" right to me, but have I missed something? Thanks for your help.
 

awkward

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Mar 2008
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Hi RogBear62,

It is not necessary to add up the various binomial probabilities to get the answer. If you re-read Mr. F's solution, you will see he gave your a simpler method.

Here is another route to the same answer, which maybe you will better understand. Let's say there are n people in the household, each of whom likes the product with probability 0.65. To find the probability that at least one person likes the product, consider the complementary problem: What is the probability that no one likes the product? This happens with probability \(\displaystyle 0.35^n\). So the probability that at least one person likes the product is

\(\displaystyle 1 - 0.35^n\).

Note that this ASSUMES that each person's likes/dislikes are statistically independent. The same assumption is necessary for Mr. F's solution. In the Real World, this is unlikely.
 
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