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kifflom [539]
4 years ago
11

You have two biased coins. Coin A comes up heads with probability 0.1. Coin B comes up heads with probability 0.6.However, you a

re not sure which is which, so you choose a coin randomly and flip it. If the flip is heads, you guess that the flipped coin is B. Otherwise, you guess that the flipped coin is A.What is the probability that your guess is correct?
Mathematics
1 answer:
Andrews [41]4 years ago
7 0

Answer:

The probability that our guess is correct = 0.857.

Step-by-step explanation:

The given question is based on A Conditional Probability with Biased Coins.

Given data:

P(Head | A) = 0.1

P(Head | B) = 0.6

<u>By using Bayes' theorem:</u>

P(B|Head) = P(Head|B) \times \frac{P(B)}{P(Head)}

We know that P(B) = 0.5 = P(A), because coins A and B are equally likely to be picked.

Now,

P(Head) = P(A) × P(head | A) + P(B) × P(Head | B)

By putting the value, we get

P(Head) = 0.5 × 0.1 + 0.5 × 0.6

P(Head) = 0.35

Now put this value in P(B|Head) = P(Head|B) \times \frac{P(B)}{P(Head)} , we get

P(B|Head) = P(Head|B) \times \frac{P(B)}{P(Head)}

P(B|Head) = 0.6 \times \frac{0.5}{0.35}

P(B|Head) = 0.857

Similarly.

P(A|Head) = 0.857

Hence, the probability that our guess is correct = 0.857.

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Consider the polynomial function.
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<h3>Rule of Descartes</h3>

This states that the number of real positive zeros of a polynomial are equal to or less than by an even number the number of sign changes of the coefficients of the polynomial, f(x). Also, the number of real negative zeros of a polynomial are equal to or less than by an even number the number of sign changes of the coefficients of the polynomial, f(-x).

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Since f(x) = x⁴ + 2x³ - 11x² - 5x - 6

The cofficients are +1, + 2, -11, -5, -6

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There is no sign change from -5 to - 6.

Since there is only one sign change, there is 1 positive zero.

<h3>The negative zero</h3>

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Learn more about rule of descartes here:

brainly.com/question/11444977

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