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What is the probability of at least one birthday match among a group of 41 people?
The probability of at least 1 match is equivalent to 1 minus the probability of there being no matches. The first person's birthday can fall on any day without a match, so the probability of no matches in a group of 1 is 365/365 = 1. The second person's birthday must also fall on a free day, the probability of which is 364/365 The probability of the third person also falling on a free day is 363/365, which we must multiply by the probability of the second person's birthday being free as this must also happen. So for a group of 3 the probability of no clashes is (363*364)/(365*365). Continuing this way, the probability of no matches in a group of 41 is (365*364*363*...326*325)/36541 This can also be written 365!/(324!*36541) Which comes to 0.09685... Therefore the probability of at least one match is 1 - 0.09685 = 0.9032 So the probability of at least one match is roughly 90%
Which term most closely matches the description An expression of possible loss adverse outcome or negative consequence such as injury or illness in terms of probability and severity?
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What is the probability of rolling a sum of 8 and doubles when rolling two dice?
The probability of rolling a sum of 8 and doubles when rolling two dice is 1 in 36, or about 0.02778. Simply note that there are 36 permutations of two dice, of which exactly one of them (a 4 and a 4) matches the conditions specified.
What is the probability that a family of 4 would have exactly 2 girls and 2 boys?
For the sake of ease, I'm going to assume that you mean: if a family were to have four children, what is the probability of having two girls and two boys. (Note: if you actually mean a family of four, you would have to start building in the probability of single-parenthood and same-sex adoption since the parents would be included in the total family size.) One could do this problem the long way by writing out all combinations of children and identifying which one meets the criteria given in the problem and which ones do not. All Possible Combos of 4 Children GGGG BGGG GBGG GGBG GGGB BBGG BGBG BGGB GBGB GBBG GGBB GBBB BGBB BBGB BBBG BBBB Each set that matches our criteria are in bold. There are 6 "successful" pairings, out of a total of 16 possible combinations. Thus the probability is 6/16 or .375 or 37.5% that there will be two girls and two boys. That is a long way of doing this problem however and would not be as quick if we had asked about 10 children instead of only 4. Instead it is easier to think of this problem as the probability of one possible combination being correct then multiplying it by all the possible ways of writing it. You can take any possible combination of B and G as long as there are 2 Bs and 2 Gs. For this problem lets use: BBGG. The probability of this exact combination occurring is 1/16: (1/2) * (1/2) * (1/2) * (1/2) = 1/16 Remember that the probability of a certain series of independent events is equal to the each probability multiplied together. Now we must figure out all the possible ways of writing BBGG. In this case there are 6. So now 6 * (1/6) = 6/16 or 3/8 or .375 or 37.5%
If you have 5 football matches with 3 outcomes win lose or draw how many different combinations are there?
3 + 3 + 3 + 3 + 3 = 15 total possible outcomes. You can 'prove' this by laying out a table of possibles where a user might tick the result of each game..... Match....1....2....3....4....5 Win......._...._...._...._...._ Lose......_...._...._...._...._ Draw....._...._...._...._...._
