anonymous
  • anonymous
MEDAL!! (refer to choosing two cards from a thoroughly shuffled deck. Assume that the deck is shuffled after a card is returned to the deck) If you put the first card back in the deck before you draw the next, what is the probability that the first card is a 10 and the second card is a jack?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jim_thompson5910
  • jim_thompson5910
Focus on one card at a time. What is the probability of pulling out a "10" card ?
jim_thompson5910
  • jim_thompson5910
here's a visual of all the cards (52 total; 4 suits, 13 per suit) http://www.jfitz.com/cards/classic-playing-cards.png
anonymous
  • anonymous
hmm, there are four 10s

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jim_thompson5910
  • jim_thompson5910
yes, so the probability of picking a single 10 is?
anonymous
  • anonymous
guessing 4 out of 52
jim_thompson5910
  • jim_thompson5910
you can reduce that fraction to get ?
jim_thompson5910
  • jim_thompson5910
4/52 = ???
anonymous
  • anonymous
1/13
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
so if you focus on one suit only, there is a 1 in 13 chance to get a "10" card
jim_thompson5910
  • jim_thompson5910
next step: find the probability of picking a jack. All 52 cards are still there since the first card was put back (it says `Assume that the deck is shuffled after a card is returned to the deck`)
anonymous
  • anonymous
if there are 4 jacks wouldn't it be 4/52=1/13 again?
jim_thompson5910
  • jim_thompson5910
yep
jim_thompson5910
  • jim_thompson5910
now multiply the two fractions
jim_thompson5910
  • jim_thompson5910
P(10 on first draw, jack on second draw) = P(10 on first draw) * P(jack on second draw) this works because the two events are independent
anonymous
  • anonymous
multiplying the two fractions would get me 1/169
jim_thompson5910
  • jim_thompson5910
which is the correct final answer
anonymous
  • anonymous
thanks again jim!

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