Question

A student studying for a vocabulary test knows the meanings of 16 words from a list of 20 words. If the test contains 10 words from the study list, what is the probability that at least 8 of the words on the test are words that the student knows

Answer 1

Answer:

$P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)$

And we can find the individual probabilites:

$P(X=8) = 10C8 (0.8)^8 (1-0.8)^{10-8}= 0.302$

$P(X=9) = 10C9 (0.8)^9 (1-0.8)^{10-9}= 0.268$

$P(X=10) = 10C10 (0.8)^9 (1-0.8)^{10-10}= 0.107$

And adding the values we got:

$P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)= 0.302+0.268+0.107= 0.678$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

For this case the probability of success is given by:

$p = \frac{16}{20}= 0.8$

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=10, p=0.8)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

And we want to find the following probability:

$P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)$

And we can find the individual probabilites:

$P(X=8) = 10C8 (0.8)^8 (1-0.8)^{10-8}= 0.302$

$P(X=9) = 10C9 (0.8)^9 (1-0.8)^{10-9}= 0.268$

$P(X=10) = 10C10 (0.8)^9 (1-0.8)^{10-10}= 0.107$

And adding the values we got:

$P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)= 0.302+0.268+0.107= 0.678$