anonymous
  • anonymous
Two cards are drawn from a standard deck of 52 cards without replacement. What is the probability that both cards are greater than 2 and less than 9?
Mathematics
schrodinger
  • schrodinger
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ParthKohli
  • ParthKohli
Okay, I'll help you with this :)
anonymous
  • anonymous
Thanks little bro, lol. I am studying and forgot this all.
ParthKohli
  • ParthKohli
What would be the probability of getting a card between 2 and 9 in the first pick?

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anonymous
  • anonymous
2/52 and 9/52
ParthKohli
  • ParthKohli
No...it's saying that it's greater than 2 and less than 9. 3,4,5,6,7,8 are the numbers that qualify for me :)
anonymous
  • anonymous
So we have to get the ones in between?
ParthKohli
  • ParthKohli
Yes.. exactly
ParthKohli
  • ParthKohli
Since there are 3,4,5,6,7,8 in four suits, the probability would be: \( \color{Black}{\Rightarrow \Large {4 \times 6 \over 52} }\) For the first pick.
ParthKohli
  • ParthKohli
Simplifying further: \( \color{Black}{\Rightarrow \Large {24 \over 52} = {12 \over 26} = {6 \over 13} }\)
ParthKohli
  • ParthKohli
Now we've picked the first. We have 1 card LESS in the deck because we haven't replaced the cards.
ParthKohli
  • ParthKohli
So this time the denominator of the probability would become 51. The numerator will also get one less because we have assumed that we have picked one card which is 3,4,5,6,7 or 8.
ParthKohli
  • ParthKohli
There are 23 cards left that we want. 51 total cards left. \( \color{Black}{\Rightarrow \Large {23 \over 51} }\)
anonymous
  • anonymous
Okay...so since we had not replaced, we have to subtract 1?
ParthKohli
  • ParthKohli
Now if we want two things to happen at the same time, we shall multiply the probabilities. \( \color{Black}{\Rightarrow \Large {13 \over 26} \times {23 \over 51}}\)
ParthKohli
  • ParthKohli
We have to subtract 1 from both numerator and denominator.
anonymous
  • anonymous
Thats it?
ParthKohli
  • ParthKohli
Nope
anonymous
  • anonymous
Wait. so after we multiply, we have to simplify further too?
ParthKohli
  • ParthKohli
Oops.. I meant this: \( \color{Black}{\Rightarrow \Large {6 \over 13} \times {23 \over 51}}\)
ParthKohli
  • ParthKohli
Multiply the fractions. The fractions are in their simplest forms so when we'll multiply it'd be in the simplest form.
anonymous
  • anonymous
Yes, i did that. Okay, so 46/221 is final answer?
ParthKohli
  • ParthKohli
How did you get that?
ParthKohli
  • ParthKohli
Oh yes
ParthKohli
  • ParthKohli
\( \color{Black}{\Rightarrow \Large {2 \over 13} \times {23 \over 17} }\)
ParthKohli
  • ParthKohli
Correct! :D
anonymous
  • anonymous
Okay, thats awesome!! Thanks soo much Parth!! Probability is my weak spot :/
ParthKohli
  • ParthKohli
Haha probability is easy...getting the hang of what it involves is important
anonymous
  • anonymous
Yea, see, im reviewing this since last semester, lol.
ParthKohli
  • ParthKohli
And I've helped someone on this site after a long time. we usually just answer questions ://
anonymous
  • anonymous
Thanks, I actually learned :D

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