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Marta_Voda [28]
3 years ago
11

An arcade booth at a county fair has a person pick a coin from two possible coins available and then toss it. If the coin chosen

lands on heads, the person gets a prize. One coin is a fair coin and one coin is a biased coin (unfair) with only a 32% chance of getting ahead. Assuming the equally likely probability of picking either coin, what is the probability that the fair coin is the one chosen, given that the chosen coin lands on heads?
a. 0.8300
b. 0.6024
c. 0.0149
d. 0.4150
e. 0.4310
Mathematics
1 answer:
nydimaria [60]3 years ago
7 0

Answer:

<em>b. 0.6024</em>

Step-by-step explanation:

<u>Conditional Probability</u>

Suppose two events A and B are not independent, i.e. they can occur simultaneously. It means there is a space where the intersection of A and B is not empty:

P(A\cap B) \neq 0

If we already know event B has occurred, we can compute the probability that event A has also occurred with the conditional probability formula

\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}

Now analyze the situation presented in the question. Let's call F to the fair coin with 50%-50% probability to get heads-tails, and U to the unfair coin with 32%-68% to get heads-tails respectively.

Since the probability to pick either coin is one half each, we have

P(U)=P(F)=50\%=0.5

If we had picked the fair coin, the probability of getting heads is 0.5 also, so

P(F\cap H)=0.5\cdot 0.5=0.25

If we had picked the unfair coin, the probability of getting heads is 0.32, so

P(U\cap H)=0.32\cdot 0.5=0.16

Being A the event of choosing the fair coin, and B the event of getting heads, then

P(B)=P(F\cap H)+P(U\cap H)=0.25+0.16=0.41

P(A\cap B)=P(F\cap H)=0.25

\displaystyle P(A|B)=\frac{0.25}{0.41}=0.61

The closest answer is

b. 0.6024

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Answer:

0.0803 = 8.03% probability that the number who have a high school degree as their highest educational level is exactly 32.

Step-by-step explanation:

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Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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P(X = 32) = C_{100,32}.(0.304)^{32}.(0.696)^{68} = 0.0803

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2 years ago
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Answer:

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Step-by-step explanation:

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²⁴³/₂₇ =

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