January 4, 200818 yr Author e. .5 X .5 X .5 = .125 Without the phrase "at least one of which is a boy, " you would be correct.
January 5, 200818 yr 1 in 7... it would be 1 in 4 if you specified WHICH child (first, second, or third) IS a boy. Allowing for any (but not all) of the three to be girls, it looks like 1 in 7.
January 5, 200818 yr Doesn't matter which is a boy. The fact is that one child is already accounted for, making the operative question "what are the chances that the other two children are boys." One in four.
January 5, 200818 yr 1/4 (1/2 times 1/2; since one is already a boy - if that hadn't already been stated, the answer would be 1/8). Good answer!:ylsuper:
January 5, 200818 yr Author The probability of any event E occurring = (number of ways E can occur)/ (number of total outcomes in the sample space). That is, P(E) = n(E)/n(S). The key to this problem is the sample space. Because of the conditional expression"at least one of which is a boy," the sample space consists of only 7 outcomes, i.e., S ={(B,B,B) , (B,B,G), (B,G,B),(B,G,G), (G,B,B) , (G,B,G), (G,G,B)}. Thus, if E is the event of three boys, then P(E) = 1 / 7 .
January 5, 200818 yr The probability of any event E occurring = (number of ways E can occur)/ (number of total outcomes in the sample space). That is, P(E) = n(E)/n(S). The key to this problem is the sample space. Because of the conditional expression"at least one of which is a boy," the sample space consists of only 7 outcomes, i.e., S ={(B,B,B) , (B,B,G), (B,G,B),(B,G,G), (G,B,B) , (G,B,G), (G,G,B)}. Thus, if E is the event of three boys, then P(E) = 1 / 7 . But you didnt specify the order in which they were born when you asked the question.. Isnt BBG the same a GBB? The outcome is still two boys and one girl. Correct?
January 7, 200818 yr Author But you didnt specify the order in which they were born when you asked the question.. Isnt BBG the same a GBB? The outcome is still two boys and one girl. Correct? No. You have to consider both outcomes.Order does matter. Consider flipping a fair coin twice. The possible outcomes are {(H,H), (H, T) , (T,H), and (T, T)} . The probability of getting "Heads" both times would be 1/4, which, of course, is correct since we know that by the multiplication rule of independent events, the probability would be 1/2 x 1/2 . If we maintained that (H, T) and (T, H) was the same outcome, then the probability of two Heads would be 1/3, which, of course, is incorrect.
January 8, 200818 yr No. You have to consider both outcomes.Order does matter. Consider flipping a fair coin twice. The possible outcomes are {(H,H), (H, T) , (T,H), and (T, T)} . The probability of getting "Heads" both times would be 1/4, which, of course, is correct since we know that by the multiplication rule of independent events, the probability would be 1/2 x 1/2 . If we maintained that (H, T) and (T, H) was the same outcome, then the probability of two Heads would be 1/3, which, of course, is incorrect. Now I know why I barely scraped by in my statistics class:lol:
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