Question

24​% of a certain​ country's voters think that it is too easy to vote in their country. You randomly select 12 likely voters. Find the probability that the number of likely voters who think that it is too easy to vote is​ (a) exactly​ three, (b) at least​ four, (c) less than eight.

Answer 1

Answer:

a) $P(X=3) = 12C3 (0.24)^3 (1-0.24)^{12-3}= 0.2573$

b)  $P(X=0) = 12C0 (0.24)^0 (1-0.24)^{12-0}= 0.037$

$P(X=1) = 12C1 (0.24)^1 (1-0.24)^{12-1}= 0.1407$

$P(X=2) = 12C2 (0.24)^2 (1-0.24)^{12-2}= 0.2573$

$P(X=3) = 12C3 (0.24)^3 (1-0.24)^{12-3}= 0.1828$

And after replace we got:

$P( X \geq 4) = 1-0.6795 = 0.3205$

c) $P(X=0) = 12C0 (0.24)^0 (1-0.24)^{12-0}= 0.037$

$P(X=1) = 12C1 (0.24)^1 (1-0.24)^{12-1}= 0.1407$

$P(X=2) = 12C2 (0.24)^2 (1-0.24)^{12-2}= 0.2573$

$P(X=3) = 12C3 (0.24)^3 (1-0.24)^{12-3}= 0.1828$

$P(X=4) = 12C4 (0.24)^4 (1-0.24)^{12-4}= 0.1828$

$P(X=5) = 12C5 (0.24)^5 (1-0.24)^{12-5}= 0.092$

$P(X=6) = 12C6 (0.24)^6 (1-0.24)^{12-6}= 0.034$

$P(X=7) = 12C7 (0.24)^6 (1-0.24)^{12-7}= 0.00921$

And we got:

$P(X$

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".  

Part a

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

$X \sim Binom(n=12, p=0.24)$  

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:  

$nCx=\frac{n!}{(n-x)! x!}$  

For this case we can use the mass function and we got:

$P(X=3) = 12C3 (0.24)^3 (1-0.24)^{12-3}= 0.2573$

Part b

For this case we want this probability:

$P(X \geq 4)$

And we can use the complement rule and we got:

$P(X \geq 4) = 1-P(X$

And we can find the individual probabilites and we got:

$P(X=0) = 12C0 (0.24)^0 (1-0.24)^{12-0}= 0.037$

$P(X=1) = 12C1 (0.24)^1 (1-0.24)^{12-1}= 0.1407$

$P(X=2) = 12C2 (0.24)^2 (1-0.24)^{12-2}= 0.2573$

$P(X=3) = 12C3 (0.24)^3 (1-0.24)^{12-3}= 0.1828$

And after replace we got:

$P( X \geq 4) = 1-0.6795 = 0.3205$

Part c

For this case we want this probability:

$P(X$

We can find the individual probabilites and we got:

$P(X=0) = 12C0 (0.24)^0 (1-0.24)^{12-0}= 0.037$

$P(X=1) = 12C1 (0.24)^1 (1-0.24)^{12-1}= 0.1407$

$P(X=2) = 12C2 (0.24)^2 (1-0.24)^{12-2}= 0.2573$

$P(X=3) = 12C3 (0.24)^3 (1-0.24)^{12-3}= 0.1828$

$P(X=4) = 12C4 (0.24)^4 (1-0.24)^{12-4}= 0.1828$

$P(X=5) = 12C5 (0.24)^5 (1-0.24)^{12-5}= 0.092$

$P(X=6) = 12C6 (0.24)^6 (1-0.24)^{12-6}= 0.034$

$P(X=7) = 12C7 (0.24)^6 (1-0.24)^{12-7}= 0.00921$

And we got:

$P(X$