i need 3 probability word problems that equal 2/3 when simplified?
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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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None. Two odds make an even
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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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This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
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In 36 possibles outcomes\nprobability of doubles\n1-1\n2-2\n3-3\n4-4\n5-5\n6-6\n6 in 36 or 1 in 6 or 16,666%\nOf the sum being 4\n1 +3\n2+2\n3+1\n3 in 36 or 1 in 12 or 8,333%\nIf it is combined probability you can sum the two\n16,66+8,33=25%
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k=1/4 or k=2 or many more possibilities
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6 x 1/3 = 2\n2 pieces.
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1. 1s2 2s2 2p6 3s1\n2. 1s2 2s2 2p6 \n3. 1s2 2s2 2p4\n4. 1s2 2s2 2p6\n5. 1s2 2s2 2p6 3s2 3p6 3d7 4s2
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try this site for help on probabiliity\n\nhttp://www.futureaccountant.com/probability/
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P(A or B) = P(A) + P(B) - P(A and B)\n\nP(A) = probability of trying out for basketball\nP(B) = probability of trying out for baseball\n\nNow, P (A and B) = 0, since Jamie will not try out for basketball and baseball\n\nP(A or B) = 1 (or 100%), since Jamie will definitely try out for basketball or baseball\n\nSo, P(A) + P(B) = 100%\nP(B) is more than 42% or more (since you have to consider that Jamie will try out for other positions for baseball).\n\n100% - 42% = 58%\nP(A) has to be 58% or lesser. The only answer that is less than 58% is option A (40)
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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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try\nhttp://www.peterwebb.co.uk/probability.htm
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There are 20 students in the class. There are 4 possibilities when you choose the 3 random students. I've included the probabilities of these 4 possibilities. 1) and 3) are the answers to your questions.\n\n1)   3 boys = (4/20)(3/19)(2/18) = 0.0035\n\n2)    2 b, 1 g = (4/20)(3/19)(16/18)x3(=3C2) = 0.0842\n\n3)    2 g, 1 b = (16/20)(15/19)(4/18)x3(=3C2) = 0.4211\n\n4)     3 g = (16/20)(15/19)(14/18) =  0.4912\n\nFor the probabilities to be correct they must sum to 1:\n\n0.0035+0.0842+0.4211+0.4912 = 1
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If X= probablity of chosing a vowel from a bag of tiles in which each tile has a letter from the word 'Mathematics' written on it..\n\nTotal letters/tiles= 11\nNo of vowels = 4\n\nX= 4/11\nX= .36
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x1*p(x1) + x2*p(x2)\n\nx1 = 1 for  event dots> 3,\nx2 = 4 for event  dots<=3.  \nThese are equally probable because there are 6 dots on the die.  \n\n1*(1/2) + 4*(1/2)
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there are 3, (since 5/3 *3 = 5)
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For that to happen, 1 must come on each dice.\n\nThus P(Sum is 2) = P(1 appears)*P(1 appears)\n\n=(1/6)^2 = 1/36