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 April 14th, 2011, 07:31 AM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Probability (confusing question) 1. The birthdays of John and Ivy are in the first seven days in January. Find the probability that next year a) Both have their birthdays on Monday b) John and Ivy have their birthday on the same day c) they have their birthdays on different day d )Monday is the birthday of one or both 2. A box containing five black ball and one white ball.Alan and Bill takes turns to draw a ball from box, starting with Alan. The first boy to draw the white ball wins the game. Assuming that they do not replace the balls as they draw them out, find the probability that Bill wins the games. If the game is changed, so that, in the new game, they replace each ball after it has been drawn out, find the probabilities that a) Alan wins at his first attempt b) Alan wins at his second attempt c) Alan wins at his third attempt Show that these answers are in a Geometric progression. Hence find the probability that Alan wins the new game. These two question are confusing me a lot. I need help, especially for question 2. Please show me your working step by step. Thank you.
April 14th, 2011, 01:08 PM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: Probability (confusing question)

Hello, hoyy1kolko!

Quote:
 1. The birthdays of John and Ivy are in the first seven days in January. Find the probability that next year: a) Both have their birthdays on Monday

$P(\text{John, Monday}) \,=\,\frac{1}{7}, \;\;\; P(\text{Ivy, Monday}) \,=\,\frac{1}{7}$

$\text{Therefore: }\;P(\text{both Monday}) \:=\:\frac{1}{7}\,\cdot\,\frac{1}{7} \:=\:\frac{1}{49}$

Quote:
 b) John and Ivy have their birthday on the same day

$\text{John's birthday can be }any\text{ day of the week: }\:P(\text{John, any day}) \:=\:\frac{7}{7} \:=\:1$

$\text{Then Ivy must have the same birthday: }\:P(\text{Ivy, same day}) \:=\:\frac{1}{7}$

$\text{Therefore: }\;P(\text{same day}) \:=\:1\,\cdot\,\frac{1}{7} \:=\:\frac{1}{7}$

Quote:
 c) They have their birthdays on different days

$P(\text{different days}) \;=\;1\,-\,P(\text{same day}) \;=\; 1 \,-\,\frac{1}{7} \;=\;\frac{6}{7}$

Quote:
 d) Monday is the birthday of one or both

$\text{The opposite of "at least one Monday" is "}neither\text{ on Monday".}$

$P(\text{neither, Monday}) \;=\;\frac{6}{7}\,\cdot\,\frac{6}{7} \;=\;\frac{36}{49}$

$\text{Therefore: }\;P(\text{at least one Monday}) \;=\;1\,-\,\frac{36}{49} \;=\;\frac{13}{49}$

 April 14th, 2011, 08:42 PM #3 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability (confusing question) Please help me to solve question 2 step by step. The question is trickier. If you can, please show me the tree diagram of the question. THX
 April 15th, 2011, 06:00 AM #4 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability (confusing question) help question 2 please .i can't solve it at all.
April 15th, 2011, 12:39 PM   #5
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: Probability (confusing question)

Hello, hoyy1kolko!

I'll explain the first part of problem #2.

Quote:
 2. A box contains five black balls and one white ball. Alan and Bill takes turns to draw a ball from box, starting with Alan. The first boy to draw the white ball wins the game. [1] Assuming that they do not replace the balls as they draw them out, find the probability that: a) Alan wins at his first attempt

$P(\text{Alan, w}) \:=\:\frac{1}{6}$

Quote:
 b) Alan wins at his second attempt

$P(\text{Alan, b}) \,\cdot\,P(\text{Bill, b})\,\cdot\,P(\text{Alan, w}) \;=\;\frac{5}{6}\,\cdot\,\frac{4}{5}\,\cdot\,\frac {1}{4} \;=\;\frac{1}{6}$

Quote:
 c) Alan wins at his third attempt

$P(\text{Alan, b})\,\cdot\,P(\text{Bill, b})\,\cdot\,P(\text{Alan, b}) \,\cdot\,P(\text{Bill, b})\,\cdot\,P(\text{Alan, w})\;=\;\frac{5}{6}\,\cdot\frac{4}{5}\,\cdot\,\fra c{3}{4}\,\cdot\,\frac{2}{3} \,\cdot\,\frac{1}{2} \;=\;\frac{1}{6}$

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Although I truly enjoy explaining Mathematics,
[color=beige]. . [/color]your lack of response is discouraging.

I post complete solutions to be used as a template for other problems.
But you don't seem to be learning from my efforts.

You continue to post problem after problem with no comments
[color=beige]. . [/color]except for "How about the second problem?"

How about "I understand now", or maybe a simple "Thank you."

 April 15th, 2011, 10:22 PM #6 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability (confusing question) oh sorry but i have learn from every method that you helped me.Thank for yr advice and thank you.

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