Question

what is the probability of getting an odd number on a fair die and a tail on a coin flip

Answer 1

@johnweldon1993

Answer 2

Hey!

So first...a fair die has 6 sides with an equal probability for each right? so the probability for any 1 number would be \(\large \dfrac16\)

And a coin has 2 sides...with equal probability again...so that chance would be \(\large \dfrac12\)

So

how many odd numbers on are a die?

Answer 3

3

Answer 4

Right...so there are 3 out of 6 numbers on a die that are off....so the chance of getting any random odd number on a die is 3/6 or 1/2 right?

SO!!

The probability of getting an odd on a die is \(\large \dfrac12\) and the probability of getting a tails on a coin is also \(\large \dfrac12\) so the probability we want for those together is

\[\large \frac{1}{2} \times \frac{1}{2} =?\]

Answer 5

btw, odd* not off* I'm sure you know that but wanted to clarify :)

Answer 6

i have another onnneee

Answer 7

what is the probability of getting a number less than three on a fair die and a tail on a coin flip

Answer 8

Well we already know the tails probability is that 1/2

So...a number LESS THAN 3

how many numbers are less than 3 on a die?

Answer 9

2

Answer 10

Indeed, so that probability is 2 numbers out of 6 ...so 2/6 or 1/3

so our total probability of the 2 events is \(\large \frac{1}{3} \times \frac{1}{2} =?\)

Answer 11

what is the probability of choosing a 2 and a face card from a deck of cards

Answer 12

Okay...standard deck of cards has 52 cards right?

How many 2's are in there?
and how many face cards are in there?

Answer 13

4 2's and 12 face cards i believe

Answer 14

mmhmm :)

4 suits and 1 2 in each so there are 4 2's
and 3 face cards in each suit so 12 face cards

SO! Now this is where I'm a little confused...if we are choosing both cards at the SAME time...we can consider these independent...

however if we choose a 2 from the deck...and THEN choose a face card from the only 51 cards remaining...that changes things...so I'm not too sure how to proceed

Answer 15

But I think...since it doesnt SPECIFY one and THEN the other...I think its just the basic choose 2 random cards at the same time...so

The probability of choosing a 2 out of 52 cards is \(\large \dfrac2{52} = \dfrac{1}{26}\)
And the probability of choosing a face card would be \(\large \dfrac{12}{52} = \dfrac{3}{13}\)

So the total probability would be \(\large \frac{1}{26} \times \frac{3}{13} = ?\)

Answer 16

3/338 which is 0.00887573964 right so the answer would be 0.00?

Answer 17

Omg....brain fart -_-

Sorry...probability of choosing a 2 would be

\[\large \frac{4}{52} = \frac{1}{13}\]

sorry about that -_- lol

so its actually

\[\large \frac{1}{13} \times \frac{3}{13}\]

Answer 18

3/169 which is 0.0177514793

Answer 19

There you go...that's better lol

as soon as I saw you put 0.00 I was like...well THAT means impossible! lol

sorry about that!

Answer 20

lol its all good

Answer 21

but my options are 0.00, 0.077, 0.25, 0.3077

Answer 22

Argh see this is why I hate probability questions that are NOT specific lol

Okay....I'm going to sum up here based on what I think

\[\large P(2) = \frac{4}{52} = \frac{1}{13}\]
\[\large P(face) = \frac{12}{52} = \frac{3}{13}\]

\[\large P(together) = P(2) + P(face) = \frac{1}{13} + \frac{3}{13} = \frac{4}{13} = 0.3077\]