Abox contains 3 coins, one fair, one weighted with a 2 3 chance of coming up heads, and one weighted with a 2 3 chance of coming up tails. you take a coin at random from the box, flip it 20 times, and get 11 heads. what is the probability you have the fair coin?

The probability that we have a fair coin is 0.2717

Step-by-step explanation:

Let Ci be the event that the coin i is used for while we let i = 1,2,3 (representing the three coin)

We also let H be the event that the coin which is flipped lands (which is head)

Therefore, the problem arise:

That:  P(H/C1) = 1/2fair coin

          p(H/C2) = 2/3Chance of head

          p(H/C3) = 2/3Chance of tail

Now, noting that the coin was picked at random,

We have, : p(Ci) = P(C1) = P(C2) = P(C3) = 1/3

We can then say that or calculate thus:

P( C2 | H ) = P (H ∩ C2) ÷ P(H)

Where P (H ∩ C2) means the probability of event intersection

= P(H ∩ C2) ÷ P(H ∩ C1) + P(H ∩ C2) + P(H ∩ C3)

= P(H | C2) P(C2) ÷ P(H | C1) P(C1) + P(H | C2) P(C2) + P(H | C3) P(C3)

= (1 / 2) (1 / 3) ÷  (1/2)(1/3) + (2/3)(1/3) + (2/3) (1/3)

0.5 × 0.3 ÷ (0.5 × 0.3) + (0.67 × 0.3) + (0.67 × 0.3)

0.15 ÷ 0.15 + 0.201 + 0.201

=0.15 / 0.552

= 0.2717

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

step-by-step explanation:

i dont understand the question? ?

step-by-step explanation: