SOLUTION: The random variable X has a binomial distribution with n = 7 and p = 0.44. Calculate P(4). Enter your answer as a decimal with four decimal places.

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Question 1197154: The random variable X has a binomial distribution with n = 7 and p = 0.44.
Calculate P(4). Enter your answer as a decimal with four decimal places.
Found 3 solutions by ikleyn, ewatrrr, math_tutor2020:
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.

The words " binomial distribution " refer you to the standard formula in your textbook.

Having the formula, you can do your calculations manually or using your calculator.

Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!

Binomial distribution: p = .44 , n=7 Find P(4) Using Formula (Recommend Using stattrek.com to check results) Using p and q are the probabilities of success and failure respectively. p = .44 , q = .56 = 35 (Using nCr button on Calculator) = .2304 Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus P(x) is binompdf(n, p, x-value). P(4) = binompdf(7, .44, 4)= .2304 Feeling comfortable with Your calculator is key to Your success working with the various Distributions.

Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!

If your teacher doesn't care to see the steps and/or time is very short, then I recommend a calculator or spreadsheet.

Here are two such examples of a free online calculator
https://www.omnicalculator.com/statistics/binomial-distribution
https://www.statology.org/binomial-distribution-calculator/

If you want to use a spreadsheet, then the command to type in is =BINOM.DIST(4,7,0.44, 0)
Alternatively you can use BINOMDIST in place of BINOM.DIST
Don't forget about the equal sign up front.
Check out this page for further documentation of how the BINOMDIST function works
https://support.microsoft.com/en-us/office/binom-dist-function-c5ae37b6-f39c-4be2-94c2-509a1480770c
This documentation page is for excel, but the function works with other spreadsheet programs like OpenOffice and Google Sheets.

On TI83 and TI84 calculators, press the button labeled "2nd" in the top left corner. Then hit the VARS key. Scroll down until you reach binompdf
and hit enter. Then type in the following
binompdf(7,0.44,4)
Unfortunately the order of these input values is slightly different from BINOMDIST mentioned earlier. Be sure to review your calculator's manual to get the correct order of values.

In GeoGebra, the command to type in is BinomialDist(7, 0.44, 4, false)

Feel free to pick your favorite calculator method.

Using any of those calculators gives an answer of 0.2304 when rounding to four decimal places.

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If you don't have access to such calculators, or your teacher won't allow it, then the standard method is to use the Binomial Distribution formula that the tutor @ewatrrr has mentioned.
That's the method I'd recommend if you are in exam settings.

But let's look at this a different way so we can see where the binomial formula comes from.

The term "binomial" means there are two outcomes.
bi = 2

Think of a coin flip.
In this case, the coin is biased toward one side.
Let's say "heads" has a probability of 0.44
That means "tails" has probability of 1-0.44 = 0.56

We'll flip the coin n = 7 times.
We could flip a single coin 7 times, or we could flip 7 coins exactly once each.
The two situations are identical.

P(4) represents the "The probability of exactly 4 heads showing up out of 7 coin tosses".
The notation is equivalent to P(x = 4).

Assuming each coin toss is independent, we multiply out the probabilities
Let's start with the heads
(0.44)(0.44)(0.44)(0.44) = (0.44)^4

Then the tails
(0.56)(0.56)(0.56) = (0.56)^3

In short,
probability of four heads = (0.44)^4
probability of three tails = (0.56)^3

Therefore (0.44)^4*(0.56)^3 represents the probability of getting the sequence HHHH TTT

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The expression (0.44)^4*(0.56)^3 applies to that sequence only.
But we could arrange the heads and tails.

For instance, we could go from this
HHHH TTT
to this
HHHT HTT

The question is: how can we count the number of ways to arrange these heads and tails?

That's where the nCr combination formula comes in.
Some textbooks will call this the "choose" formula.

Imagine slips of paper labeled 1 through 7. Then place those pieces of paper into a hat.
We'll randomly select 4 slips of paper.

One possible sequence is 1,2,5,7
Those numbers will then tell us where to place the "H"s to get this sequence of coin flips: H,H,_,_,H,_,H
I've filled slots 1,2,5 and 7.

Each black space will then get a T to get HHTT HTH

The process of selecting those 4 pieces of paper can be counted by using the nCr combination formula.

We have n = 7 pieces of paper and r = 4 selections
Let's compute the nCr value
n C r = (n!)/(r!(n-r)!)
7 C 4 = (7!)/(4!*(7-4)!)
7 C 4 = (7!)/(4!*3!)
7 C 4 = (7*6*5*4!)/(4!*3!)
7 C 4 = (7*6*5)/(3!)
7 C 4 = (7*6*5)/(3*2*1)
7 C 4 = (210)/(6)
7 C 4 = 35
Put another way, we have 7*6*5 = 210 permutations, and then we divide by 3*2*1 = 6 to adjust for the fact that order doesn't matter.

So that's where the 35 comes from in the equation @ewatrrr mentioned.

You can also use Pascal's Triangle to determine this. Locate the row that has 1,7,... at the start. Then count 4+1 = 5 spaces to arrive at 35.

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Recall that (0.44)^4*(0.56)^3 represented the probability of getting the sequence HHHH TTT.

Then we counted 35 ways to arrange the four H's and three T's.

This means we'll stick 35 out front to get
35*(0.44)^4*(0.56)^3

Computing that expression will get roughly 0.2303789694976 which then rounds to 0.2304 which we got earlier.