<p>I'm assuming we can talk about it now. The question that was like, if you choose an integer from one to 10 twice with replacement, what is the chance you get greater than 5, that was .9 right?</p>
<p>I believe I did get .9 for that.</p>
<p>Analogy: throwing two 10-sided dice and getting at least one of the numbers in the 6-10 range.
75 favorable out of 100 possible outcomes.
.75</p>
<p>Or: chances of getting no numbers at all from the 6-10 range
.5 * .5 = .25.
So 1 - .25 = .75</p>
<p>wasn't it, the sum of the two numbers greater than 5?
so 11, 12, 21, 13, 31, 14, 41, 23, 32, 22 <below five.
(100-10)/100= .9??</p>
<p>I wasn't sure how to do it, so I just did a few simulations on the calc and it came pretty close to .9 each time.</p>
<p>I brute forced this one.</p>
<p>If 1 is one of the numbers, then 5, 6, 7, 8, 9, and 10 would work as the other number. That's 6/10.</p>
<p>If 2 is one of the numbers, then 4, 5, 6, 7, 8, 9, and 10 would work. That's 7/10.</p>
<p>If 3 is one of the numbers, then 3, 4, 5, 6, 7, 8, 9, and 10 would work. That's 8/10.</p>
<p>If 4 is one of the numbers, then 2, 3, 4, 5, 6, 7, 8, 9, 10 would work. That's 9/10.</p>
<p>5 and on, all other numbers would work. 5-10 are all 10/10.</p>
<p>6+7+8+9+10+10+10+10+10+10/100 = .9</p>
<p>i guess gcf didn't take the test. since it was the sum of the two numbers.
Arti, you did all those work for a question....lol</p>
<p>haha well it was one of the ones I skipped and came back to later. i'm so bad at probability; luckily the ones on SATs are simple enough</p>