Two dice are rolled. Let E be the event that the sum of the outcomes is odd and F be the event of at least one 1. Interpret the events EF, EcF, and EcFc.
Q:Two dice are rolled. Let E be the event that the sum of the outcomes is odd and F be the event of at c c cleast one 1. Interpret the events EF, E F, and E F .AnswerStep 1 of 1(a)The sample space of an experiment consists of all possible outcomes of the experiment.We have given two dice are rolled, then all possible outcomes are as follows, (1,1)(1,2)(1,3)(1,4)(1,5)(1,6) (2,1)(2,2)(2,3)(2,4)(2,5)(2,6) (3,1)(3,2)(3,3)(3,4)(3,5)(3,6) (4,1)(4,2)(4,3)(4,4)(4,5)(4,6) (5,1)(5,2)(5,3)(5,4)(5,5)(5,6) (6,1)(6,2)(6,3)(6,4)(6,5)(6,6)Let E be the event that the sum of the outcomes is odd.Therefore, E = (1,2)(1,4)(1,6) = (2,1)(2,3)(2,5) = (3,2)(3,4)(3,6)\n = (4,1)(4,3)(4,5) = (5,2)(5,4)(5,6) = (6,1)(6,3)(6,5)Let F be the event of at least one.Hence, F = (1,1)(1,2)(1,3)(1,4)(1,5)(1,6) = (2,1)(3,1)(4,1)(5,1)(6,1)We need to findEF or the intersection of both events or common outcomes from both theevents,Hence, EF = E F = common outcomes EF = E F = (1, 2) Hence EF is(1, 2). cWe need to findE F. cE = complimentary event of E = the outcomes which is not there in the sample space of E\n c E = sum of the outcomes is even E = (1,1)(1,3)(1,5) = (2,2)(2,4)(2,6) = (3,1)(3,3)(3,5) = (4,2)(4,4)(4,6) = (5,1)(5,3)(5,5) = (6,2)(6,4)(6,6) F = (1,1)(1,2)(1,3)(1,4)(1,5)(1,6) = (2,1)(3,1)(4,1)(5,1)(6,1) Hence, c c E F = E F = common outcomes c c E F = E F = (1, 1) c Hence E F is (1, 1). c cWe need to find E F . c cF = complimentary event of F = the outcomes which is not there in the sample space of F F c = (2,2)(2,3)(2,4)(2,5)(2,6)\n = (3,2)(3,3)(3,4)(3,5)(3,6) = (4,2)(4,3)(4,4)(4,5)(4,6) = (5,2)(5,3)(5,4)(5,5)(5,6) = (6,2)(6,3)(6,4)(6,5)(6,6) c E = (1,1)(1,3)(1,5) = (2,2)(2,4)(2,6) = (3,1)(3,3)(3,5) = (4,2)(4,4)(4,6) = (5,1)(5,3)(5,5) = (6,2)(6,4)(6,6)Hence, E F = E F = common outcomes c c c c E F = E F = (2,2)(2,4)(2,6)(4,2)(4,4)(4,6)(6,2)(6,4)(6,6) = (3,3)(3,5)(5,3)(5,5)
