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One of two coins is selected at random and tossed three times. The first coin comes up heads with probability P₁ -1/3 and the second coin with probability P₂-2/3. (a) What is the probability that the mber of heads is k-3 (b) Repeat (a) for k-0,1,2 d) In part (c), which coin is more probably when 2 heads have been observed?

User Valerij
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Final answer:

The probability of getting a certain number of heads when a coin is selected at random and tossed three times can be calculated using the binomial probability formula. The probabilities for getting 3, 0, 1, and 2 heads are -1/27, 64/27, -32/27, and 4/9 respectively.

Step-by-step explanation:

To find the probability of getting a certain number of heads when a coin is selected at random and tossed three times, we need to consider the probabilities of each case. Let's denote P₁ as the probability of getting a head with the first coin (-1/3) and P₂ as the probability of getting a head with the second coin (-2/3).

(a) To find the probability of getting exactly k=3 heads, we can use the binomial probability formula: P(k) = (3 choose k) * P₁^k * (1-P₁)^(3-k), where (n choose k) is the binomial coefficient. Substituting values, P(3) = (3 choose 3) * (-1/3)^3 * (1-(-1/3))^(3-3) = 1 * (-1/3)^3 * (4/3)^0 = -1/27.

(b) Similarly, we can calculate the probabilities for k=0, 1, and 2 heads using the same formula. P(0) = (3 choose 0) * (-1/3)^0 * (1-(-1/3))^(3-0) = 1 * 1 * (4/3)^3 = 64/27. P(1) = (3 choose 1) * (-1/3)^1 * (1-(-1/3))^(3-1) = 3 * (-1/3) * (4/3)^2 = -32/27. P(2) = (3 choose 2) * (-1/3)^2 * (1-(-1/3))^(3-2) = 3 * (1/9) * (4/3)^1 = 4/9.

Based on these calculations, the probabilities are: P(3) = -1/27, P(0) = 64/27, P(1) = -32/27, P(2) = 4/9.

User Amru
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