Probability of getting a head? on to Probability || fall20There are three coins in a box, one fair coin, one coin with heads on both sides and one weighted coin with a1/3 probability of getting a head on each toss. If we randomly select and then flip a coin. What is theprobability of getting a head?Select one:О a. 6/18O b. 3/18Ос 10/18O d. 11/18Next page

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College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter8: Sequences, Series, And Probability

Section8.7: Probability

Problem 37E

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Probability of getting a head?

on to Probability || fall20
There are three coins in a box, one fair coin, one coin with heads on both sides and one weighted coin with a
1/3 probability of getting a head on each toss. If we randomly select and then flip a coin. What is the
probability of getting a head?
Select one:
О a. 6/18
O b. 3/18
Ос 10/18
O d. 11/18
Next page

Expert Solution

Step 1

Given that the there are three coins in a box.

$L e t C_{1} : u n b i a s e d c o i n C_{2} : t w o h e a d e d c o i n C_{3} : b i a s e d c o i n$

$H : h e a d a p p e a r s o n t h e c o i n$

$P (C_{1})$ $=$ Probability that coin selected is unbiased = $\frac{1}{3}$

$P (C_{2}) =$ Probability that coin selected is two headed = $\frac{1}{3}$

$P (C_{3}) =$ Probability that coin selected is biased coin = $\frac{1}{3}$

Step 2

$P (H / C_{1}) = P r o b a b i l i t y o f h e a d a p p e a r o n t h e c o i n C_{1} = \frac{1}{2} P (H / C_{2}) = P r o b a b i l i t y o f h e a d a p p e a r o n t h e c o i n C_{2} = 1 P (H / C_{3}) = P r o b a b i l i t y o f h e a d a p p e a r o n t h e c o i n C_{3} = \frac{1}{3}$

We need to find P(H).

By using Law of Total Probability , we get

$P (H) = P (C_{1}) * P (H / C_{1}) + P (C_{2}) * P (H / C_{2}) + P (C_{3}) * P (H / C_{3})$

$P (H) = \frac{1}{3} * \frac{1}{2} + \frac{1}{3} * 1 + \frac{1}{3} * \frac{1}{3} = \frac{1}{6} + \frac{1}{3} + \frac{1}{9} = \frac{(3 + 6 + 2)}{18} = \frac{11}{18}$

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