# And rule for multiple selections

• MHB
• Irish teacher
In summary, the product rule for probabilities holds if the probabilities of A and B are not independent.
Irish teacher
Hello all I'm new as I'm just revising my maths to teach irish leavers in revising and exams maths and sciences. This exam section has me puzzled hope you can help

Three digit numbers are made from the numbers 2,3,4,5,6 .. See questions and my calls below

Total number of outcomes is ... 5x4x3 = 60

Number of outcomes under 400 ... 2/5 x 60 = 24
Number of outcomes divisible by 5 ... 1/5 x 60 = 12

Here's the tricky bit
Number of outcomes both under 400 and divisible by five... If I list these there are 6 but if I try the AND equation it's 4.8 ( 1/5x2/5 x 60) what am I missing in the calculation?

I'm sure it's something obvious but I'm completely blind to it. Hope you can help IT

Irish teacher said:
Hello all I'm new as I'm just revising my maths to teach irish leavers in revising and exams maths and sciences. This exam section has me puzzled hope you can help

Three digit numbers are made from the numbers 2,3,4,5,6 .. See questions and my calls below

Total number of outcomes is ... 5x4x3 = 60

Number of outcomes under 400 ... 2/5 x 60 = 24
Number of outcomes divisible by 5 ... 1/5 x 60 = 12

Here's the tricky bit
Number of outcomes both under 400 and divisible by five... If I list these there are 6 but if I try the AND equation it's 4.8 ( 1/5x2/5 x 60) what am I missing in the calculation?

I'm sure it's something obvious but I'm completely blind to it. Hope you can help IT

Hi Irish teacher! Welcome to MHB!

Which AND rule are you talking about?

Alternatively, we have:

Number of outcomes under 400 ...
First digit must be 2 or 3: 2 possibilities.
Second digit can be any of the 4 remaining digits.
Third digit can be any of the 3 remaining digits.
So in total: 2 x 4 x 3 = 24

Number of outcomes both under 400 and divisible by five...
First digit must be 2 or 3: 2 possibilities.
Last digit must be 5: 1 possibility.
Second digit can be any of the 3 remaining digits.
So in total: 2 x 3 x 1 = 6.

Hi ILS thankyou so much for your replyThat's so useful.I'm concerned though as we teach the AND and OR equations So if it asks below 400 or divisible by 5 we would say
OR

P(aor b) = pa + pb -( p (a+b).

Then if it's 400 and divisible by 5 we would use the AND equation
P( a+b) = pa x pb

Which clearly doesn't work in this case. I guess it's about identifying which questions require these and which is looking for the type of solution above. Thanks again for your reply.

IT

I like Serena said:
Hi Irish teacher! Welcome to MHB!

Which AND rule are you talking about?

Alternatively, we have:

Number of outcomes under 400 ...
First digit must be 2 or 3: 2 possibilities.
Second digit can be any of the 4 remaining digits.
Third digit can be any of the 3 remaining digits.
So in total: 2 x 4 x 3 = 24

Number of outcomes both under 400 and divisible by five...
First digit must be 2 or 3: 2 possibilities.
Last digit must be 5: 1 possibility.
Second digit can be any of the 3 remaining digits.
So in total: 2 x 3 x 1 = 6.

The product rule for probabilities is $\DeclareMathOperator\and{and} P(A\and B)=P(A)P(B)$, which holds only if A and B are independent.
In our case, 'less than 400' and 'divisible by 5' are unfortunately not independent.

So we would need to apply the general product rule for probabilities: $P(A\and B)=P(A\mid B)P(B)$.
Then if we draw a random number with those digits, we have:
$$P(\text{divisible by 5}\and \text{below 400}) =P(\text{divisible by 5} \mid\text{below 400})\,P(\text{below 400}) \\ =\frac{\text{# divisible by 5 given first digit is below 4}}{\text{# numbers below 400}}\cdot\frac{\text{# numbers below 400}}{\text{#total}} =\frac{2\cdot 3\cdot 1}{2\cdot 4\cdot 3}\cdot \frac{2\cdot 4\cdot 3}{5\cdot 4\cdot 3} = \frac 6{60}$$

## What is the "And rule" for multiple selections?

The "And rule" is a principle in statistics that states that when multiple independent events must occur together, the probability of all events occurring is equal to the product of their individual probabilities.

## How is the "And rule" used in scientific research?

The "And rule" is commonly used in research to calculate the probability of multiple independent variables occurring together. This is important for making accurate predictions and drawing conclusions from data.

## What are some examples of the "And rule" in action?

Some examples of the "And rule" in action include predicting the likelihood of a specific combination of genetic traits occurring in offspring, calculating the probability of multiple risk factors leading to a disease, and determining the chances of a particular combination of weather conditions occurring.

## Are there any limitations to the "And rule"?

Yes, the "And rule" assumes that the events being considered are truly independent of each other. In reality, there may be hidden connections or factors that can affect the outcomes, making the calculated probabilities less accurate.

## How is the "And rule" different from the "Or rule"?

The "And rule" and the "Or rule" are two different principles used in statistics to calculate the probability of multiple events occurring. The "Or rule" states that the probability of at least one of multiple events occurring is equal to the sum of their individual probabilities, while the "And rule" calculates the probability of all events occurring together.

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