A car park holds 100 cars of many different colours. How can the number of yellow cars still be binomial?

Because the trial is not what colour a car is, but whether it is yellow , and that question has exactly two outcomes: yellow, or not yellow. Defining the trial as a yes or no question about each item is what reduces any number of categories to two, and it is how most binomial models are set up. The other conditions still have to hold before the model can be used.

Someone eats 5 sweets from a bag holding 6 caramels and 4 marshmallows, and wants to model the number of caramels eaten . Which condition fails?

The probability of success is not constant , and the trials are not independent . Eating a caramel first leaves 5 caramels among 9 sweets rather than 6 among 10, so the probability changes with every sweet taken and depends on what has already been eaten.

30% of a city's population has blue eyes and a sample of 30 people is taken. What lets this be treated as binomial?

That the population is large and the sample is random , which together let each person be treated as having a constant 0.3 probability of blue eyes, independently of the others. Strictly the trials are not independent, since once a blue-eyed person has been picked there is one fewer left to pick, but in a large population removing 30 people barely changes the proportion. So the binomial is being used here as a good approximation , not as an exact description.

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Binomial Distribution (AQA AS Maths: Statistics): Flashcards

Exam code: 7356

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FrontThe Binomial Distribution
What four conditions must be satisfied before a binomial distribution can be used?
BackThe Binomial Distribution
There must be a fixed number of trials, $n$ , and:
each trial is independent of the others
each trial has exactly two outcomes, success or failure
the probability of success, $p$ , is constant
If any one of them fails the model does not apply, and independence is worth checking first, because it is the condition that most often fails in a real situation.
BackThe Binomial Distribution
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Cards in this collection (12)

What four conditions must be satisfied before a binomial distribution can be used?
There must be a fixed number of trials, $n$ , and:
- each trial is independent of the others
- each trial has exactly two outcomes, success or failure
- the probability of success, $p$ , is constant
If any one of them fails the model does not apply, and independence is worth checking first, because it is the condition that most often fails in a real situation.
Complete the formula for the probability of exactly $x$ successes in $n$ trials:
$P (X = x) = (\binom{n}{x}) p^{______} {(1 - p)}^{______}$
The completed formula is:
$P (X = x) = (\binom{n}{x}) p^{x} {(1 - p)}^{n - x}$
If $x$ of the $n$ trials are successes then the remaining $n - x$ must be failures, which is where the two indices come from.
The coefficient $(\binom{n}{x})$ counts the number of different orders in which those $x$ successes could occur, since the formula has to cover every one of them.
True or False?
The distributions $B (10, 0.2)$ and $B (10, 0.8)$ are mirror images of each other.
True.
Replacing $p$ by $1 - p$ swaps the roles of success and failure, so counting successes in one is the same as counting failures in the other.
$B (10, 0.2)$ peaks at 2 with its tail to the right, $B (10, 0.8)$ peaks at 8 with its tail to the left, and each probability in one reappears in the other at $10 - x$ .
$X ~ B (n, p)$ has mean 225 and variance 144. How do you find $p$ ?
Divide the variance by the mean, because $\frac{n p (1 - p)}{n p} = 1 - p$ and the $n p$ cancels.
Here $\frac{144}{225} = 0.64$ , so $1 - p = 0.64$ and $p = 0.36$ .
The mean of a binomial is $n p$ and its variance is $n p (1 - p)$ , so their ratio strips out $n$ altogether and leaves the probability of failure.
A car park holds 100 cars of many different colours. How can the number of yellow cars still be binomial?
Because the trial is not what colour a car is, but whether it is yellow, and that question has exactly two outcomes: yellow, or not yellow.
Defining the trial as a yes or no question about each item is what reduces any number of categories to two, and it is how most binomial models are set up.
The other conditions still have to hold before the model can be used.
Someone eats 5 sweets from a bag holding 6 caramels and 4 marshmallows, and wants to model the number of caramels eaten. Which condition fails?
The probability of success is not constant, and the trials are not independent.
Eating a caramel first leaves 5 caramels among 9 sweets rather than 6 among 10, so the probability changes with every sweet taken and depends on what has already been eaten.
30% of a city's population has blue eyes and a sample of 30 people is taken. What lets this be treated as binomial?
That the population is large and the sample is random, which together let each person be treated as having a constant 0.3 probability of blue eyes, independently of the others.
Strictly the trials are not independent, since once a blue-eyed person has been picked there is one fewer left to pick, but in a large population removing 30 people barely changes the proportion.
So the binomial is being used here as a good approximation, not as an exact description.
Which calculator function gives $P (X = x)$ , and which gives $P (X \leq x)$ ?
The Binomial Probability Distribution function, often shortened to Binomial PD or Bpd, gives $P (X = x)$ , the probability of exactly $x$ successes.
The Binomial Cumulative Distribution function, Binomial CD or Bcd, gives $P (X \leq x)$ , the probability of $x$ successes or fewer.
Both take the same three inputs, the value of $x$ , the number of trials $n$ and the probability of success $p$ , so picking the wrong one returns a plausible but wrong answer.
Complete the identity for $P (X \geq x)$ on a calculator that only gives $P (X \leq x)$ :
$P (X \geq x) = 1 - P (X \leq______)$
The completed identity is:
$P (X \geq x) = 1 - P (X \leq x - 1)$
The values left out of $X \geq x$ are those below $x$ , and the largest of them is $x - 1$ , so that is where the cumulative function has to stop.
So $P (X \geq 10) = 1 - P (X \leq 9)$ , and using $P (X \leq 10)$ instead would wrongly remove $X = 10$ as well.
$X ~ B (40, 0.35)$ . How do you find $P (8 < X < 15)$ ?
Start by listing which whole numbers are wanted: strictly between 8 and 15 means 9 to 14.
Then take everything up to 14 and remove everything up to 8:
$P (8 < X < 15) = P (X \leq 14) - P (X \leq 8) = 0.542$
Writing down the wanted integers first is the reliable way to get both cumulative values right, whatever mixture of strict and weak inequalities the question happens to use.
True or False?
For $X ~ B (n, p)$ , $P (X < 5)$ and $P (X \leq 4)$ are the same.
True.
A binomial variable counts successes, so it can only take whole-number values, and the only whole numbers below 5 are 0, 1, 2, 3 and 4.
That is why any strict inequality can be rewritten as a weak one before a cumulative function is reached for, with $P (X > 5)$ becoming $P (X \geq 6)$ in the same way.
Your calculator's cumulative binomial function accepts a lower and an upper bound. What do you enter for $P (X \geq x)$ ?
Enter $x$ as the lower bound and $n$ as the upper bound, since $n$ successes is the most that $n$ trials can produce.
No identity is needed then, because the function adds the probabilities from $x$ up to $n$ directly.

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