<p>If Pr(A)=0.25, Pr(B)=0.6, and Pr(union of A and B)=0.7, what is the value of Pr (intersection of A and B)?..didn’t understand the problem</p>
<p>The first box contains 7 blue balls and 2 red balls. the 2nd box contains 6 blue balls and 8 red balls. If a ball is choose randomly from each box, what is the probability that at leat one ball is red…annoying probabilities.</p>
<p>from SAT practices…doing those once a week…</p>
<p>You can just subtract the probability of none being red from 1.
For the first box, prob of no red is 7/9. Second- 3/7. Multiply these to get the prob of no red. Then subtract from 1.</p>
<p>Draw the Venn diagrams for A and B; call it a black oval for A, a white oval for B, with a color of gray for the overlapping area. You need the probability of the ‘gray area’.</p>
<p>P(B only) = P(A union B) - P(A)
= 0.7 - 0.25 = 0.45
P(B) = P(B only) + P(A intersection B)
0.6 = 0.45 + x
So x = P(A intersection B) = 0.6 - 0.45 = 0.15</p>
<p>The ‘gray’ area has P=0.15, the ‘black’ area is P(A) - 0.15 = 0.10, and the ‘white’ area is P(B) - 0.15 = 0.45 .
(You can do it without the Venn diagrams, but I find it helps me to visualize the problem)</p>
<p>Probability that you’ll pick a red ball from box 1 = 2/9
probability that you’ll pick a red ball from box 2 = 4/7
Probability that you will pick either a red ball from box one OR a red ball from box 2 = 2/9 + 4/7 = 50/63</p>
<p>someone correct me if i’m wrong.</p>
<p>I’m not sure caa5042, but I think your solution leaves out the possibility of both balls being red.</p>
<p>yeah- thought about that</p>
<p>probabilty of both balls being red = 2/9 * 4/7 = 8/63</p>
<p>8/63 + 50/63 = 58/63</p>
<p>i think your answer was 42/63 though. So i’m not positive.</p>
<p>Which exam are these questions from?</p>
<p>Caa5042:
(WriteOnTheTransom had the correct answer = 1 - (7/9)(3/7) = 1 - 1/3 = 2/3 )</p>
<p>In your logic, you are double-counting the probability of two reds. It is a much longer way of getting the answer, but if you insist :), it should be</p>
<p>Prob(at least 1 red) = Prob(1 red from box1 and 0 red from box 2)
+ Prob( 0 red from box1 and 1 red from box 2)
+ Prob( 1 red from box1 and 1 red from box 2)
= (2/9)(6/14) + (7/9)(8/14) + (2/9)(8/14)
= 6/63 + 28/63 + 8/63 = 42/63 = 2/3 .</p>
<p>For the union question the formula is this: </p>
<p>p(A intersection B)= p(a)+p(b)- p(a union b). </p>
<p>Since the question gives you the p (a), the p(b), and the p(a U b), the p(a intersection b)= .25+.6-.7= 0.15. </p>
<p>I have no idea about the red ball one.</p>
<p>caa5042, by that logic, flip two coins and you’re guaranteed that one will be heads 
(1/2 + 1/2 = 1)</p>
<p>Both writonthetransom’s and optimizerdad’s logics are correct.</p>
<p>instead of finding the probability of not picking any red balls and subtracting from one, is there an equally simple way of finding the answer straight away by using the respective probabilities of picking one red ball from the first and second boxes?</p>
<p>why is it not p(1 blue from box 1)<em>p(1 red from box 2) + p(1 red from box 1)</em>p(1 blue from box 2) + p(1 red from box 1)*p(1 red from box 2)</p>
<p>is 7/9<em>8/14 + 2/9</em>6/14 + 2/9*8/14 = 2/3</p>
<p>that IS the answer.Same as 42/63</p>
<p>1)</p>
<p>0.25+0.6-0.7=0.15</p>
<p>2)</p>
<p>2/3</p>
<p>optimizer- never mind. Schoenhs explanation has made it clear now. The ‘0’ red thing in your explanation threw me off- I didn’t notice that ‘0’ red means 1 blue.</p>
<p>oh, oops, i didn’t read the posts, guess i should have.</p>