Question

A financial talk show host claims to have a 55.3 % success rate in his investment recommendations. You collect some data over the next few weeks, and find that out 10 recommendations, he was correct 3 times. If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successfu

Answer 1

Answer:

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

Step-by-step explanation:

For each recommendation, there are only two possible outcomes. Either it was a success, or it was a failure. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$p = 0.553, n = 10$

If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successful:

This is

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.553)^{0}.(0.447)^{10} = 0.0003$

$P(X = 1) = C_{10,1}.(0.553)^{1}.(0.447)^{9} = 0.0039$

$P(X = 2) = C_{10,2}.(0.553)^{2}.(0.447)^{8} = 0.0219$

$P(X = 3) = C_{10,3}.(0.553)^{3}.(0.447)^{7} = 0.0724$

$P(X = 4) = C_{10,4}.(0.553)^{4}.(0.447)^{6} = 0.1567$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0003 + 0.0039 + 0.0219 + 0.0724 + 0.1567 = 0.2552$

There is a 25.52% probability of observating 4 our fewer succesful recommendations.