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Probability Tree DiagramsDate: 8/22/96 at 1:20:30 From: Philip Carter Subject: Probabilities I wonder if you would help to settle an arguement by providing answers to the three problems below: 1. A woman has two children. What are the odds that both are boys? 2. Charlie hits the target 80 times in 100 shots. Jim hits the target 90 times in 100 shots. What are the chances that the target will be hit if each fires once? 3. In 1946, statistics showed that 2 percent of fruit boats arrived with their cargoes ruined. If two boats arrived, what was the probability that both cargoes were ruined? Philip Carter
Date: 8/22/96 at 2:36:22
From: Doctor Paul
Subject: Re: Probabilities
Let's set up a tree diagram:
50% / \ 50%
boy girl
50% / \50% 50%/ \50%
boy girl boy girl
To find the odds, just multiply across the path that leads to two
boys:
.50 * .50 = .25 or 25 percent.
Had you wanted to know the odds of a boy and a girl, you would have
multiplied across the path that leads to a boy and a girl and then add
that to the path that leads to a girl and then a boy. They both
satisfy the criteria so you add them.
2. Again, let's set up a tree diagram...assume that Charlie goes
first.
80%/ \20%
hits misses
90%/ \10% 90%/ \10%
hits misses hits misses
Let's see which branches lead to the target being hit:
if it is hit and then hit again, that counts; if it is hit and then
missed, that counts; if it is missed and then hit, that counts; if
both miss it, that doesn't count... so let's multiply along the first
three branches and then add:
(.8 * .9) + (.8 * .1) + (.2 * .9) = .72 + .08 + .18 = .98,
or 98 percent. Pretty good odds, eh?
3. Tree diagram again... (are you noticing a theme?)
98%/ \2%
good bad
98%/ \2% 98%/ \2%
good bad good bad
Both ruined: .02 * .02 = .0004 = .04 percent (pretty low)
I hope this helps you out. These problems are easy if you use tree
diagrams!
Regards,
-Doctor Paul, The Math Forum
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