#$&* course MTH 152 2/20 5:30 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
.............................................
Given Solution: `athere are 3 possible odd outcomes and four outcomes less that 5 which would add up to 7 outcomes, except that 2 of the outcomes < 5 are alrealdy odd and won't be counted. Thus the number of outcomes which are odd or less that 5 is 3 + 4 - 2 = 5 (this expresses the rule that n(A U B) = n(A) + n(B) - n(A ^ B), where U and ^ stand for union and intersection, respectively ). Thus the probability is 5/6. In terms of the specific sample space: The sample space for the experiment is {1, 2, 3, 4, 5, 6}. Success corresponds to events in the subset {1, 2, 3, 4, 5}. There are 6 elements in the sample space, 5 in the subset consisting of successful outcomes. Thus the probability is 5/6. ** Self-critique OK ------------------------------------------------ Self-critique Rating: OK ********************************************* question: Query 12.2.15 drawing neither heart nor 7 from full deck YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 36 possible ways to not draw a hear nor a 7 from a full deck of cards. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThe sample space consists of the 52 cards in a full deck. There are 39 cards that aren't hearts, four 7's but only three aren't hearts so there are 36 cards that aren't hearts or seven. The probability is therefore 36/52 = 9/13. The odds in favor of the event are 16 to 36 (number favorable to number unfavorable), which in reduced form is 4 to 9. ** Self-critique OK ------------------------------------------------ Self-critique Rating: OK ********************************************* question: 12.2.24 prob of black flush or two pairs YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 123,552 possible ways to get two pairs from a full deck of cards and 2,574 ways to get a black flush. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThere are C(13,5) = 1287 ways to get a flush in a given suit--gotta choose the 5 cards from the 13 cards in that suit. There are two black suits so there are 2 * 1287 = 2574 possible black flushes. As the text tells you there are 123,552 ways to get two pairs. You can incidentally get this as 13 * C(4, 2) * 12 * C(4, 2) * C(44, 1) / 2 (2 of the 4 cards in any of the 13 denominations, then 2 of the 4 cards in any of the remaining 12 denominations, divide by 2 because the two denominations could occur in any order, then 1 of the 44 remaining cards not in either of the two denominations. There is no way that a hand can be both a black flush and two pairs, so there is no overlap to worry about (i.e., n(A and B) = 0 so n(A or B) = n(A) + n(B) - n(A and B) = n(A) + n(B) ). Thus there are 123,552 + 2574 = 126,126 ways to get one or the other. The probability is therefore 126,126 / 2,598,960 = .0485, approx. ** Self-critique OK ------------------------------------------------ Self-critique Rating: OK