My Math Forum Two Probability Questions, Stumped!

 December 13th, 2012, 07:22 PM #1 Newbie   Joined: Dec 2012 Posts: 4 Thanks: 0 Two Probability Questions, Stumped! So I have these two probability questions and probability is my weakest subject. These are suppose to be simple but I cannot wrap my mind around it. The first problem: "Suppose a family wants to have 3 children. Find the probability that the family has at least 1 boy." The second problem: "If a marble is drawn from a bag containing 4 yellow, 5 red, and 3 blue marbles, what are probabilities of the following? a. The marble is blue. b. The marble is blue or red" Help would be appreciated! Thanks
 December 13th, 2012, 07:42 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Two Probability Questions, Stumped! 1.) It is certain given the couple has 3 children, that they will either have 3 girls (event A) OR at least 1 boy (event B). So we may state: $P(A)+P(B)=1$ $P(B)=1-P(A)=1-$$\frac{1}{2}$$^3=\frac{7}{8}$ I am assuming the probability of having either gender is equal. 2.) There are 12 marbles total. a) There are 3 blue marbles, so $P(X)=\frac{3}{12}=\frac{1}{4}$ b) There are a total of 8 blue or red marbles, so $P(X)=\frac{8}{12}=\frac{2}{3}$
 December 15th, 2012, 05:22 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond Re: Two Probability Questions, Stumped! #1: B B B * B B G * B G B * G B B * G G G G G B * G B G * B G G * P(B) = 7/8.

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