Probability problems

Hello guys, I need help with the following problems:

1. select a positive 2 digit number less than 50. P(factor of 4)
2. select a positive 2 digit number greater than 59. P(factor of 7)
3. select a factor of 72. P(factor will be odd)
4. toss a coin 3 times. P(not getting 2 tails and 1 head)
5. toss 3 coins. P(getting 1 tail and 2 heads)
6. a coin is tossed 7 times. 3 times it comes out heads and 4 times it comes out tails. What is the probability that you will get tails on the next loss?

Thanks in advance!!

These are simple counting problems. You should be practicing freely. If you are struggling with these, truly struggling, you may wish to rethink your class schedule. Seriously.

"select a positive 2 digit number less than 50."

10, 11, 12, ...49 -- How many are there in that list? Make sure "05" isn't considered a 2-digit number.

"P(factor of 4)"

Words really are important. I am almost certain you mean MULTIPLE of 4. The FACTORS of 4, being 1, 2, and 4, don't produce very many 2-digit numbers.

12, 16, 20. ... 48 -- How many are there in that list.

The challenge is four fold:

1) Understand the problem statement.
2) Count the members in the population.
3) Count the members of interest.
4) Accurately perform whatever arithmetic is required.

I know these are extremely easy problems. What I don't understand is this:

in the first problem it says "select a positive 2 digit number less than 50. P(factor of 4)"

ok, because you said it has to be multiple, not factors, I counted all the 2 digit multiples of 4. (10)

so --------------> P(multiple of 4) has to be 10/100 ??

I don't know what to put on the denominator. Do I have to put only the 2 digit numbers? (91) or all the numbers from 1-100 (100)

is the answer this 10/91 or this 10/100


thanks

Where did you get 100? Let's try again.

"select a positive 2 digit number less than 50."

10, 11, 12, ...49 -- How many are there in that list? Make sure "05" isn't considered a 2-digit number.

megadeth95 said:

What I don't understand is this:
in the first problem it says "select a positive 2 digit number less than 50. P(factor of 4)"
ok, because you said it has to be multiple, not factors, I counted all the 2 digit multiples of 4. (10)

Click to expand...

\(\displaystyle \dfrac{99}{4}=24.75\) so there are 24 positive integers less that 100 that are divisible by 4. But that includes 4, 8 which are not two-digit numbers.
There are ninety two-digit numbers less than 100.
So what is the answer?

Ok, this is the problem:

"select a positive 2 digit number less than 50. P(multiple of 4)"

these are the multiples of 4: 12,16,20,24,28,32,36,40,44,48 = 10 in total

However, there are 41 2-digit numbers from 1 to 50 (10,11,12, ...50)

therefore the answer is 10/41

is it correct? thx in advance

megadeth95 said:

Ok, this is the problem:
"select a positive 2 digit number less than 50. P(multiple of 4)"
these are the multiples of 4: 12,16,20,24,28,32,36,40,44,48 = 10 in total
However, there are 41 2-digit numbers from 1 to 50 (10,11,12, ...50)
therefore the answer is 10/41

Click to expand...

Why do you say there are 41 two-digit numbers less than 50?
There are 49 positive integers less than 50.
The numbers 1,2,3,4,5,6,7,8,9 are not two-digit integers.
SO?

the reason why I say 41 is because in the question says "select a positive 2 digit number less than 50"

It's only asking for 2 digit numbers, that is, 41. It doesn't say 1 digit numbers (1,2,3...9) or, do I have to include the 1 digit numbers so the answer will be 10/49?????

Thank you