SOLUTION: The probability if winning a prize in game of chance is 0.48. The least number of games that must be played to ensure that the probability of winning is at least eight prize is mor

Question 1203205: The probability if winning a prize in game of chance is 0.48. The least number of games that must be played to ensure that the probability of winning is at least eight prize is more than 0.95 is closest to:
A. 5
B. 10
C. 20
D. 24
E. 25
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13216) (Show Source): You can put this solution on YOUR website!

Look at what you are posting before you post it.

As posted, it is impossible to know what the question is.

"The least number of games that must be played to ensure that the probability of winning is at least eight prize is more than 0.95 is closest to..."

???

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

The portion "The least number of games that must be played to ensure that the probability of winning is at least eight prize is more than 0.95 is closest to" seems strangely worded.

I'm assuming it should read "The least number of games that must be played to ensure that the probability of winning at least eight prizes is more than 0.95 is closest to"

Rephrased another way: We want to know how many games to play so that is more than 0.95
We also want this value of n to be the smallest possible.

n = number of games played
x = number of wins
p = probability of winning = 0.48

We have a binomial probability distribution because...
There are two outcomes: you either win or you lose. Hence the "bi" in "binomial".
Each trial (aka each game played) is independent of any other.
The probability of winning any particular game is 0.48 regardless how many games are played.
Open up your favorite spreadsheet program.
The BINOMDIST function will calculate the cumulative binomial probabilities that we want.

The template for that function is
BINOMDIST(x, n, p, c)
where
x = number of successes
n = number of trials
p = probability of success
c = cumulative flag (ie it tells the spreadsheet whether we want cumulative or not)
We'll use c = 1 to turn on the cumulative flag.
p = 0.48 mentioned earlier
The phrasing "at least eight" means "eight or more".
The BINOMDIST function calculates the cumulative binomial probabilities from x = 0 to x = the current value.
If we go from x = 0 to x = 7, then take the complement of this, then we'll have the interval
So therefore, we'll involve x = 7

The value of n is unknown. We'll use the spreadsheet software to list the choices
n = 5
n = 10
n = 20
n = 24
n = 25
Let's list those values without the "n = " in front.
Place those values in column A.
5 goes in A1, 10 goes in A2, etc.

Move to cell B1
Type =BINOMDIST(7,A1,0.48,1) into this cell.
Do not forget about the equal sign up front. Otherwise the command won't execute.

The result of this calculation is an error (the specific error code will depend on which spreadsheet app you have).
Why is this? Well notice how x = 7 is larger than n = 5.
We cannot possibly have more successes than trials.
We need to have be the case.

In cell B2, we'll have =BINOMDIST(7,A2,0.48,1) as the input.
The result is the approximate value 0.95803
We don't get an error here because we have fewer successes (x = 7) compared to trials (n = 10).

You might be thinking that we found the answer since we got something larger than 0.95
But hold on. The BINOMDIST cumulative feature computes the sum of P(0)+P(1)+P(2)+...+P(x)
What the 0.95803 represents is the area to the left of x = 7, and not to the right.

approximately when n = 10.

This is why we need to subtract it from 1.
1 - 0.95803 = 0.04197

So approximately when we have n = 10 trials.

Through similar steps, you should have approximately when there are n = 20 trials.

For n = 24, approximately
For n = 25, approximately

We have crossed over the 0.95 threshold when reaching n = 24.
As n gets larger, the value will get closer to 1. It won't reach 1 itself.
Recall that

Answer: 24 (choice D)

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