Firstly, we can convert all of the fractions into percentages. To do this, we need to make the denominator of the fraction 100, and whatever we do to the denominator we must also do to the numerator.

5 x 20 = 100
1 x 20 = 20.

So Carl recieves 20/100 or 20% of the votes.

4 x 25 = 100
1 x 15 = 25

So Conroy receives 25/100 or 25% of the votes.

If we add these together and Gilda's 5%, we get 50%. Since there are 100% votes overall, we need to do 100 - 50 = 50.

Kyla receives 50% of the votes.

8 0

3 years ago

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Suppose that you are given a bag containing n unbiased coins. You are told that n-1 of these coins are normal, with heads on one

gladu [14]

Answer:

The (conditional) probability that the coin you chose is the fake coin is 2/(1 + n)

Step-by-step explanation:

Given

Total unbiased coin = n

Normal coins =n - 1

Fake = 1

The (conditional) probability that the coin you chose is the fake coin is represented by

P(Fake | Head)

And it's calculated as follows;

P(Fake | Head) = P(Fake, Head) ÷ P(Head) ----- (1)

Where P(Fake, Head) = P(Fake) * P(Head | Fake)

P(Fake) = 1/n --- because only one is fake

P(Head | Fake) = n/n because all coins (including the fake) have head

So, P(Fake, Head) = P(Fake) * P(Head | Fake) becomes

P(Fake, Head) = 1/n * n/n

P(Fake, Head) = 1/n

P(Head) is calculated by

P(Fake) * P(Head | Fake) + P(Normal) * P(Head | Normal)

P(Fake) * P(Head | Fake) = P(Fake, Head) = 1/n (as calculated above)

P(Normal) * P(Head | Normal) = ½ * (n - 1)/n ----- considering that the coin also has a tail with equal probability as that of the head.

Going back to (1)

P(Fake | Head) = P(Fake, Head) ÷ P(Head) becomes

P(Fake | Head) = (1/n) ÷ ((1/n) + (½(n-1)/n))

= (1/n) ÷ ((1/n) + (½(n-1)/n))

= (1/n) ÷ (1/n + (n - 1)/2n)

= (1/n) ÷ (2 + n - 1)/(2n)

= (1/n) ÷ (1 + n)/(2n)

= (1/n) * (2n)/(1 + n)

= 2/(1 + n)

Hence, the (conditional) probability that the coin you chose is the fake coin is 2/(1 + n)

5 0

3 years ago