Answered: You have in your pocket three $2 coins…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Related questions

Question

You have in your pocket three $2 coins and four
$1 coins. You choose three at random. What is the
probability that you then have $5 or more in hand?

Expert Solution

Step 1: Probabilities obtained.

3 coins are chosen at random one by one from $2 and $1,the possible sample space S is

S={(1,1,1),(1,1,2),(1,2,1),2,1,1),(1,2,2),(2,2,1),(2,1,2),(2,2,2)}

A={the sum is 5} ={(1,2,2),(2,1,2),(2,2,1)}

B={the sun is 6} ={(2,2,2)}

P9we have more than 5 in hand) P(A)+P(B) = $3 over 8 plus 1 over 8 equals 4 over 8 equals 0.5$

However if they are not chosen one by one ,then :

Let X be the value of $2 coin and Y be the value of $1 coin then using hypergeometric distribution

P(Sum is 5) = P(X=4,Y=1) = $fraction numerator open parentheses table row 6 row 4 end table close parentheses open parentheses table row 4 row 1 end table close parentheses over denominator open parentheses table row 10 row 5 end table close parentheses end fraction$ = $fraction numerator 15 x 4 over denominator 252 end fraction equals space 0.2381$

P(Sum is 6) = P(X=6,Y=0) = $fraction numerator open parentheses table row 6 row 6 end table close parentheses open parentheses table row 4 row 0 end table close parentheses over denominator open parentheses table row 10 row 6 end table close parentheses end fraction equals 1 over 210 space equals 0.0048$

P(Sum $greater or equal than$ 5) =0.2381+0.0048 =0.2429