<p>A deck of 52 cards contains four suits. Each suit contains 13 cards, numbered 1 through 13. Jonah picks two cards at random from the deck, without replacing the first card in the deck before drawing the second. What is the probability that the sum of the two cards he draws will be 3?</p>
<p>a) 2/13
b) 2/156
c) 2/169
d) 4/663
e) 8/663</p>
<p>I'm the worst at probability :(</p>
<p>Probability is all the same: desired amount / total amount.</p>
<p>But first, you need to figure out how many positive numbers add up to 3. The only ones are 1 and 2. Also, it just asks for the probability of the sum, not that the cards need to be picked in a specific order.</p>
<p>Cards have four suits (diamond, hearts, spades, and clubs). You want to pick a 1 or a 2 on your first draw, but it doesn’t matter which one, meaning your desired amount for the first draw should be 8/52. Simplify that to 2/13.</p>
<p>Then, they ask for no replacement, and consider that you picked a 1 from the first draw. Now, this part is tricky, because there are 4 ones, and 4 twos. Remove all the ones because 1 + 1 /= 3. Then, subtract the 4 cards you are taking out to bring your total card count to 48. Now, you should have the numbers you want for the second draw: 4/48 or 1/12.</p>
<p>Lastly, multiply the two results and your answer should be 2/156. However, that can be simplified so it is weird that the answer isn’t.</p>
<p>I got 8/663.</p>
<p>Since the only combinations that work are 1 and 2, and 2 and 1.</p>
<p>The chances of getting 1, then 2 are. (4/52)*(4/51) = 4/663.</p>
<p>Chances of 2 and 1 are the same.</p>
<p>Add them together and you get 8/663 .</p>
<p>I believe there is a fallacy in your argument :)</p>
<p>The answer is 8/663</p>
<p>x+y=3</p>
<p>The domain for the variables x and y are all integers on [1,13]. </p>
<p>The only combinations that work are (1, 2) or (2, 1). </p>
<p>When you first pick a card out of the deck, you have an 8/52 chance of picking either a 1 or a 2. (1 of hearts, spades, clubs, diamonds or 2 of hearts, spades, clubs, diamonds). If this card is picked, you will have a remaining total of 51 cards and 3 x’s and 4 y’s left (assuming x was picked). However, since you picked x first, you need to pick y second, not x again (if you picked 1, you now need a 2 and vice versa). There are still 4 y’s left, so your chance of picking a y is 4/51. To find the ultimate probability, we multiply the probabilities of both events happening together. </p>
<p>(8/52) * (4/51) = 8/663</p>
<p>wedgedawg, stop doing sat :P</p>
<p>be content with your 2400</p>
<p>This is how I distract myself from homework now. I feel guilty if I’m not doing something constructive at all times (so no video games, no TV, and reading on weekends only or right before bed). If I start doing something ‘fun’, that little voice in my head starts screaming YOU COULD BE WRITING COLLEGE ESSAYS. </p>
<p>;_;</p>
<p>Whoops I would’ve gotten that wrong! I’m sorry! Now I’m just spewing false information lol.</p>
<p>Oh okay I removed all the ones, and I was supposed to only remove one. Well that’s what I get at 1am!</p>
<p>Thank you WedgeDawg & gensis!! That helped a lot :)</p>