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 28% chance of getting a head. assuming 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?

The probability that the fair coin is the one chosen, given that the chosen coin lands on heads, is 0.64.

Step-by-step explanation:

Let H denote the event of coin toss getting a head. Let "Fair" and "Biased" refer to choices of the fair and the biased coins being tossed. P(A|B) refers to the conditional probability of event A given event B.

We are to determine the conditional probability P(Fair|H). This can be rewritten using the Bayes rule as:

in which P(Fair)=0.5 and P(H|Fair)=0.5. The only quantity that needs further thought is P(H), namely the a-priori probability of heads. This is a result of heads from a fair coin and a biased coin:

Using this value, the desired probability becomes:

The probability that the fair coin is the one chosen, given that the chosen coin lands on heads, is 0.64.

connie < felicia < stew

step-by-step explanation:

stew:

we need to find the slope

m = (y2-y1)/ (x2-x1)

    = (2650-795)/(10-3)

    = 1855/7

265 words per minute

connie:

we need to find the slope

m = (y2-y1)/ (x2-x1)

we have two points (0,0) and (4,800)

m = (800-0)/(4-0)

  m = 800/4

    200 words per minute

felicia:

260 words per minute

200< 260 < 265

connie < felicia < stew

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb