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yes..you can look at it that way too....probability of A OR B where a and b are independent events...actually i like your way better...

hey sorry for the confusion ok i made a calculation mistake for the unbiased coin problem its actually 7/36 1 + 3 3 + 1 2 + 2 2+ 3 3 + 2 1 + 4 4 + 1 7 desired outcomes and 6 * 6 = 36 total possible outcomes, so 7/36, and this is slightly greater than cards (which is 1/6). So yeah B is the answer which is greater than the probaility for card

You sent me a PM and accused me of plagiarism...can you point me to the specific post where i 'plagiarized'?

ok..but the official answer is B? shouldnt they be equal? Cards : probability is 1/6 Dice: so its 6*6 = 36 total outcomes and 2 + 2, 3 + 1, 1+3, 1 + 4, 4+ 1, 2+3 , 3+ 2 (6 desired outcomes) so 6/ 36 or 1/6

hey sorry i made a mistake.... the probability of getting a 4 or 5 on the first dice is 2/6 = 1/3 the probability of getting 4 or 5 on second dice is also= 1/3 probability of getting 4 or 5 on dice 1 AND 4 or 5 on dice 2 is = 1/9 1/3 + 1/3 - 1/9 = 2/3 - 1/9 = 5/9? God whats the answer ? lol .

i think for the two dice the probability is P(4 or 5) = P(4) + P(5) - P(4)*P(5) = 1/6 + 1/6 - (1/6 * 1/6) = 2/6 - 1/36 = 11/36 (almost 1/3 but a little bit less) what is the correct answer?

lol i hate this problem

no they dont have the same speed . but it takes the same amount of time for biker to travel to the front and return to the back, as it takes for the battalion to travel 1 KM. i think the answer should be 12 km/hr

hey steffin i want it!!! where can i find it?

speed of the battalion is 6 km/hr and they have traveled 1 km during the bikers round trip so the time elapsed during the whole process is 6 km/hr = 1 km / time time = 1/6 hr (this is also the time it takes for the biker to come to the front of the battalion and return to the back) so to find the speed of the biker, you know he traveled a total of 2 km (1 km to the front and 1 km back) so its speed = distance/ time speed = 2km / (1/6 hr) = 12 km /hr ?? whats the answer?

wow..im not taking the GMAT (taking the gre) but seems like GMAT math is A LOT harder than gre math anyway let me give it a try so the number is 40% greater than x, whish is x + 0.4x 1.4 x and its equal to 20% less than y which is 0.8y so 1.4x = 0.8 y y/ x = 7/ 4 ratio of y to x is 7:4 need to figure out what percent increase in 4 will give you 7 4 + (a/100)*4 = 7 a/100 = 3/4 a = (3/4)* 100 = 75%

i think masuhara is right. if you have 5 different colors then there are 5 C 2 = 5!/ (3! 2!)= 10 ways to pick two different colors.So that takes care of 10 distribution centers. Then you can use the 5 colors for the rest of the 5 distribution centers So there'll be 10 distribution centers with two colors and 5 with 1 color each you cant do it with 4 or less colors this is a very hard problem, by the way lol ..

interesting....have to keep that in mind....

oh wow...now i see why...this a VERY tricky question if you think of a box whose length is 6, width is 5 and height is 4, then only six boxes can fit at the bottom, and an additional six on top. You really have to draw the actual box and compare the length and width with the dimensons of the cube to see why 15 wont fit... where's this problem from? this has me a little nervous

hey the newest edition. for the revised GRE. i think its the 19th edition