anonymous
  • anonymous
what is the probability of getting an odd number on a fair die and a tail on a coin flip
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
johnweldon1993
  • 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?
anonymous
  • anonymous
3

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johnweldon1993
  • 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 \[\large \frac{1}{2} \times \frac{1}{2} =?\]
johnweldon1993
  • johnweldon1993
btw, odd* not off* I'm sure you know that but wanted to clarify :)
anonymous
  • anonymous
i have another onnneee
anonymous
  • anonymous
what is the probability of getting a number less than three on a fair die and a tail on a coin flip
johnweldon1993
  • 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?
anonymous
  • anonymous
2
johnweldon1993
  • 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 \(\large \frac{1}{3} \times \frac{1}{2} =?\)
anonymous
  • anonymous
what is the probability of choosing a 2 and a face card from a deck of cards
johnweldon1993
  • 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?
anonymous
  • anonymous
4 2's and 12 face cards i believe
johnweldon1993
  • 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
johnweldon1993
  • 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 \(\large \frac{1}{26} \times \frac{3}{13} = ?\)
anonymous
  • anonymous
3/338 which is 0.00887573964 right so the answer would be 0.00?
johnweldon1993
  • 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 \[\large \frac{1}{13} \times \frac{3}{13}\]
anonymous
  • anonymous
3/169 which is 0.0177514793
johnweldon1993
  • 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!
anonymous
  • anonymous
lol its all good
anonymous
  • anonymous
but my options are 0.00, 0.077, 0.25, 0.3077
johnweldon1993
  • 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 \[\large P(2) = \frac{4}{52} = \frac{1}{13}\] \[\large P(face) = \frac{12}{52} = \frac{3}{13}\] \[\large P(together) = P(2) + P(face) = \frac{1}{13} + \frac{3}{13} = \frac{4}{13} = 0.3077\]

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