My Math Forum University Stats - Simple Probability :)

 September 23rd, 2013, 11:36 AM #1 Newbie   Joined: Sep 2013 Posts: 4 Thanks: 0 University Stats - Simple Probability :) Hi everyone, if someone can help me with this question/explain it it'd be helpful! Consider a family with 3 children. Assume the probability that one child is a boy is 0.5 and the probability that one child is a girl is also 0.5, and that the events "boy" and "girl" are independent. a) what is the proability that all 3 children are male? notice that the complacement of the event "all three children are male" is "at least one of the children is female" use this information to compute the probabilty that at least one child is female also- i thought the complacement meant the opposite of what they are trying to achieve, so wouldn't that be all 3 children are female? THANKS!
 September 23rd, 2013, 11:58 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: University Stats - Simple Probability :) It is certain that either all three children are male OR at least 1 child is a girl...this is what makes the two events complements. To find the probability that all three children are boys, we need to consider the probability that the first child is a boy AND the second child is boy AND the third child is a boy, which gives us: $P$$\text{All 3 boys}$$=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2 }=\frac{1}{8}$ Now, because we know: $P$$\text{All 3 boys}$$+P$$\text{At least 1 girl}$$=1$ We may arrange this as: $P$$\text{At least 1 girl}$$=1-P$$\text{All 3 boys}$$$ And using our previous result, we find: $P$$\text{At least 1 girl}$$=1-\frac{1}{8}=\frac{7}{8}$ This is computationally simpler than using the binomial probability: $P$$\text{At least 1 girl}$$=P$$\text{1 girl}$$+P$$\text{2 girls}$$+P$$\text{All 3 girls}$$={3 \choose 1}$$\frac{1}{2}$$$$1-\frac{1}{2}$$^2+{3 \choose 2}$$\frac{1}{2}$$^2$$1-\frac{1}{2}$$+{3 \choose 3}$$\frac{1}{2}$$^3=\frac{3}{8}+\frac{3}{8}+\frac{ 1}{8}=\frac{7}{8}$ Whenever I see the words "at least one" in a probability problem, I take this as a cue to use the complements rule to greatly simplify matters, particularly when there are a large number of events. For three events, it's not too complicated to use the binomial formula, but for a larger number of events, it can be quite cumbersome, whereas the complement rule gets you there rather quickly.
 September 23rd, 2013, 12:36 PM #3 Newbie   Joined: Sep 2013 Posts: 4 Thanks: 0 Re: University Stats - Simple Probability :) you are fantastic! thank you so much!

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