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vegashikingdude
 #1 Posted : Thursday, February 16, 2012 3:03:58 AM
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My first day here so I'm going to post here my first 2 questions:

1. If a family has one girl and plans to have another child, answer the following if the probability of a girl is 1/2:

a) What is the probability of the 2nd child's being a boy?

b) What is the probability of the family having one girl and one boy given their current family?

2. On a roulette wheel are 36 slots numbered 1-36 and 2 slots 0 and 00. You can bet a single number. If the balls lands on your number, you receive 35 chips plus the chips you played

a) What is the probability that the ball will land on 17?

b) What are the odds against the ball landing on 17?

c) If each chip is worth \$1, what is the E.V for a player who plays the number 17 for a long time?

lisa
 #2 Posted : Monday, February 20, 2012 1:23:48 PM
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Joined: 3/16/2011
Posts: 33

[vegashikingdude wrote:]

My first day here so I'm going to post here my first 2 questions:

1. If a family has one girl and plans to have another child, answer the following if the probability of a girl is 1/2:

a) What is the probability of the 2nd child's being a boy?

b) What is the probability of the family having one girl and one boy given their current family?

2. On a roulette wheel are 36 slots numbered 1-36 and 2 slots 0 and 00. You can bet a single number. If the balls lands on your number, you receive 35 chips plus the chips you played

a) What is the probability that the ball will land on 17?

b) What are the odds against the ball landing on 17?

c) If each chip is worth \$1, what is the E.V for a player who plays the number 17 for a long time?

Hi! I will try my best to help you out :)

1a) You can either get a boy or a girl. If the probability to get a girl is 1/2 then the probability to get a boy i 1-(1/2)=1/2

1b) I'm not sure what you're suppose to calculate. They already have one girl according to the facts given. The probability that the next child is a boy is 1/2 så that would be the probability. But maybe that's not what they're asking for?

2a) To calculate the probability you need to know how many possible outcomes you have and how many favorable outcomes you have,

Number of possible outcomes = 36 (1-36) + 2 (0 and 00)=38

Number of favorable outcomes = 1 (#17)

P(17) = 1/38 = 0,026 = 2,6%

2b) To calculate the probability that the ball will not end up on number 17 we do the same as in 2a. We calculate the number of possible outcomes and the number of favorable outcomes

NUmber of possible outcomes = 38

Number of favorable outcomes = 37 (everything except 17)

P(not 17)=37/38=0,974=97,4%

2c) Not sure what E.V. stands for. Can you specify so that I do not lead you wrong :)

talouv
 #3 Posted : Thursday, March 22, 2012 1:21:47 AM
Rank: Newbie

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Joined: 3/22/2012
Posts: 2

2b) could also be solved by realizing that the (X not equal 17) is the cumplementary to (X equal 17). Then we have that

P(not 17) = 1 - P(17) = 100% - 2.6% = 97.4%.

This method is usally quiet usefull, especially when we aren't in a discrete case.

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