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

1. anonymous

@johnweldon1993

2. johnweldon1993

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?

3. anonymous

3

4. johnweldon1993

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 $\large \frac{1}{2} \times \frac{1}{2} =?$

5. johnweldon1993

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

6. anonymous

i have another onnneee

7. anonymous

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

8. johnweldon1993

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?

9. anonymous

2

10. johnweldon1993

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 $$\large \frac{1}{3} \times \frac{1}{2} =?$$

11. anonymous

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

12. johnweldon1993

Okay...standard deck of cards has 52 cards right? How many 2's are in there? and how many face cards are in there?

13. anonymous

4 2's and 12 face cards i believe

14. johnweldon1993

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

15. johnweldon1993

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 $$\large \frac{1}{26} \times \frac{3}{13} = ?$$

16. anonymous

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

17. johnweldon1993

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 $\large \frac{1}{13} \times \frac{3}{13}$

18. anonymous

3/169 which is 0.0177514793

19. johnweldon1993

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!

20. anonymous

lol its all good

21. anonymous

but my options are 0.00, 0.077, 0.25, 0.3077

22. johnweldon1993

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 $\large P(2) = \frac{4}{52} = \frac{1}{13}$ $\large P(face) = \frac{12}{52} = \frac{3}{13}$ $\large P(together) = P(2) + P(face) = \frac{1}{13} + \frac{3}{13} = \frac{4}{13} = 0.3077$