i need 3 probability word problems that equal 1/6 when simplified?
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Matthew rolled four dice. What is the probability of the total of the spots equaling 14? The answer is 1/6.\n\nJenna decided to repeat Matthew's experiment, except she rolled six dice. What is the probability of the total of the spots equaling 21? The answer is 1/6.\n\nLarry decided to repeat Jenna's experiment, but with ten dice this time. What is the probability of the total of the spots equaling 35? The answer is 1/6.\n\nGive a formula as to show when 2n dice are rolled, which total of dice will have a probability of 1/6 of being rolled.
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5| 2 -9 -7 13\n..... 10 ..5 -10\n-----------------\n....2 1 ..-2 3\n\nSo P(5) = 3
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a) probability of rolling a 2 is 1 out 6. for 5 balanced dice\n(1/6)(1/6)(1/6)(1/6)(1/6) = (1/7776)\n\nb) isnt this the same as part a?
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Well, you have to do one for each case.\n3 pos. and 2 neg. would leave 0 imaginary\n3 pos. and 0 neg. would leave 2 imaginary.\n1 pos. and 2 neg. -- 2 imaginary\n1 pos. and 0 neg. -- 4 imaginary\nAll the possible case have been covered.
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first you figure out the number of possible outcomes you can have. which would be 6x6 = 36. \nThen you figure out how many combinations you can make to make it greater than 7:\n5 and 3\n5 and 4\n5 and 5\n4 and 4\n4 and 5\n3 and 5\nwhich is 6.\n6 over 36 is one sixth. therefore your answer is 1/6.
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1) this operation is made up of two steps, that means the probablility of the whole operation = the multiplication of the probablility of every step\n a) probaility of a 1 from the first box = 1/6, the other box = 1/7 , so P(Both will be 1's) = 1/7 * 1/6\n\nb) p(Exactly one of the will be a 2) = 1/6 * 6/7 + 1/7 * 5/6 \n\nc) P(Both will be even) = 3/6 * 3/7 \nd) P( Exactly one of them will be even) = 3/6 *4/7 + 3/6 * 3/7\n\ne) P(At least one of them will be even) = 1- both of the odd = 1- (3/6 * 4/7)\n\n>> as for the sample space it would take me forever to write it down! sorry\n\n2) Not sure about this one\n3) a) 5%\n    b) 16% >> dont know wut details to include its just a law
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1) Find the probability of getting a red ball when a bag contains 2 red balls and one blue ball.\n\n2) There is a red ball, a blue ball and a green ball in a bag. Find the probability of getting either a red ball or a blue ball.\n\n3) There are 10 yellow chips and 20 purple chips in a sack. When one chip is selected at random, find the probability of NOT getting yellow chips.
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.................... / ....... &#92;\n................ 0.85 ..... 0.15 (Pass test 1)\n.............. / ....... &#92;\n.......... 0.85 ..... 0.15 (Pass test 2)\n........ / ....... &#92;\n... 0.85 ..... 0.15 (Pass test 3)\n(Fail all 3)\n\n1) How many outcomes does your tree show?\n\nThere are four outcomes:\nP(Pass Test 1) = 0.15 = 15%\nP(Pass Test 2) = 0.85 * 0.15 = 12.75%\nP(Pass Test 3) = 0.85 * 0.85 * 0.15 = 10.8375%\nP(Fail all 3) = 0.85 * 0.85 * 0.85 = 61.4125%\n\n2) What is the probability she will eventually pass an exam.  \n\nThis is the sum of all outcomes where she passes:\nP(Pass Test 1) + P(Pass Test 2) + P(Pass Test 3)\n= 38.5875%\n\nAlternatively you could take the 1 - P(Fails all 3) and get the same answer.\n\n3) What is the probability she will take the exam three times?\n\nIt didn't say whether she passes or fails, so the chance she has to take the test a third time is 0.85 * 0.85 = 72.25%\n\n4) What is the expected number of times she will take the exam?\n15% of the time, she'll take it once\n12.75% of the time, she'll take it twice\n72.25% of the time, she'll take it three times.\n\nSo the expected number of times is:\n0.15 * 1 + 0.1275 * 2 + 0.7225 * 3 = 2.5725\n\nSo she's expected to take it 2.5725 times.\n\n5) What is the expected number of times she will fail the exam?\n15% of the time she will pass test 1 (failures = 0)\n12.75% of the time she will pass test 2 (failures = 1)\n10.8375% of the time she will pass test 3 (failures = 2)\n61.4125% of the time she will fail all 3 (failures = 3)\n\n0.15 x 0 + 0.1275 x 1 + 0.108375 x 2 + 0.614125 x 3 = 2.186625\n\nSo she's expected to fail it 2.186625 times.
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A.  The probability that it will BE solved is 1 - prob that it WON'T be solved.  That is 1.00 - (1.00 - .015 [Tom doesn't]) * (1.00 - 0.50 [Alice doesn't]).  \n\n = 1.00 - (.85) * (.50).  = 1.000 - .425 = .575\n\nB.  If Tom solves it then it is 150 / 575 = 6/23
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First get it straight. Algebra and probability are 2 very different things! Now:\n\n1. There are 12 marbels, so,\na) purple then red = (1/12)*(3/12)=1/48\nb) red then green = (3/12)*(3/12) = 1/16\nc) 2 greens = (3/12)*(3/12) = 1/16\nd) 2 reds = (3/12)*(3/12) = 1/16\ne) purple then green = (1/12)*(3/12)=1/48\nf) red then yellow = (3/12)*(5/12) = 5/48\n\n2. There are 9 total marbles, so:\na) purple then white = (3/9)*(1/8) = 1/24\nb) white then green = ((1/9)*(1/8) = 1/72\nc) 2 purples = (3/9)*(2/8) = 1/12\nd) 2 yellow = (4/9)*(3/8) = 1/6\ne) yellow then purple = (4/9)*(3/8) = 1/6\nf)yellow then purple = (4/9)*(3/8) = 1/6\ng) green then white = (1/9)*(1/8) = 1/72\n\n13. Red card is 1/2 and club is 1/4, add them = 3/4\n14. Dimond is 1/4 ans spade is 1/4, add them = 1/2\n15. face card is 12/52 and a 2 is 4/52, add them = 16/52\n16. A 7 is 4/52 and a 9 is 4/52, add them = 8/52\n17. Red card is 26/52 and a king of spades is 1/52 = 27/52\n18. Heart is 13/52 and quuen od diamonds is 1/52 = 14/52
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Blue: 15/50 (simplified to 3 out of 10)\nyellow: 8/50 (simplified to 4/25)\nRed: 27/50
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1) 1, 2, 3 and 6\n2) 1, 2, 2m\n3) 1\n4) 1, 3\n\n5) 5/9 (55/99) > (45/99) 5/11\n6) 2/3 (8/12) < (9/12) 3/4\n7) 4/5 (32/40) < (35/40) 7/8\n8) 1/3 (4/12) = (4/12) 4/12
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Binomial distribution, (a) p=0.5, n=12; (b) p=0.5, n=13.  You can do it!
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P(Last is W) = P{(2nd is W and 1st is W) or (2nd is B and 1st is W)}\n= P(2nd is W and 1st is W) + P(2nd is B and 1st is W) (since these are mutually exclusive events)\n= P(2nd is W)P(1st is W|2nd is W) + P(2nd is B)P(1st is W|2nd is B) (both use the general multiplication rule)\n= (6/10)(7/11) + (4/10)(6/11)\n= (42+24)/110\n= 66/110\n= 0.6\n\nWhich is the probability of getting a white from either box, which I think makes sense, since both have the same proportion of whites.
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for these types of problems. all u have to do is change around the words to symbols, and then u got ur equation\n\njust remember "is" is the same thing as =\njust remember "of" is the same thing as * or times (multiplication)\n\n0.06 is what percent of 250\n0.06 = n% times 250\n (divide 0.06 by 250)\nsolve\n\ndo the same thing for the other problem
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There are 6 combinations\n3,5,7\n3,7,5\n5,3,7\n5,7,3\n7,5,3\n7,3,5
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sample space {1,2,3,4,5,6}