Agambler has a fair coin and a two-headed coin in his pocket. he selects one of the coins at random. (a) when he flips the coin, what is the probability that it will show heads? (b) the coin shows heads. now what is the probability that it is the fair coin?

a) probability of choosing heads= 1/2 (50%)

b)  probability of choosing the fair coin knowing that it showed heads is= 1/3 (33.33%)

Step-by-step explanation:

Since the unfair coin can have 2 heads or 2 tails , and assuming both are equally possible . then

probability of choosing the fair coin  (named A)= 1/2

probability of choosing an unfair coin with 2 heads (named B)= (1-1/2)*1/2= 1/4

probability of choosing an unfair coin with 2 tails (named C)= (1-1/2)*(1-1/2)= 1/4

then

probability of choosing heads= probability of choosing A * probability of getting heads from A + probability of choosing B * probability of getting heads from B + probability of choosing C * probability of getting heads from C =

1/2*1/2 + 1/4*1 + 1/4*0 = 2/4 = 1/2

the probability of choosing the fair coin knowing that it showed heads is

P(A/B) = P(A∩B)/P(B)

denoting event A= the coin is fair and event B= the result is heads

P(A∩B) = 1/2*1/2 = 1/4

but since we know now that that the unfair coin is not possible , the probability of choosing heads is altered:

P(B)=probability of choosing heads= probability of choosing A * probability of getting heads from A + probability of choosing B * probability of getting heads from B  = 1/2*1/2+1/2*1 = 3/4

then

P(A/B) = P(A∩B)/P(B)  = (1/4)/(3/4) = 1/3

then the probability is 1/3

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

hope to and hope that my answer is brainliest.(follow me )

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im not sure