MHB And rule for multiple selections

[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

 

[I like Serena]

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.

 

[Irish teacher]

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

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.

 

[I like Serena]

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:
$$P(\text{divisible by 5}\and \text{below 400})
=P(\text{divisible by 5} \mid\text{below 400})\,P(\text{below 400}) \\
=\frac{\text{# divisible by 5 given first digit is below 4}}{\text{# numbers below 400}}\cdot\frac{\text{# numbers below 400}}{\text{#total}}
=\frac{2\cdot 3\cdot 1}{2\cdot 4\cdot 3}\cdot \frac{2\cdot 4\cdot 3}{5\cdot 4\cdot 3}
= \frac 6{60}
$$