An envelope contains three cards: a black card that is black on both sides, a white card that is white on both sides, and a mixe
1 answer:
Answer:
There is a 2/3 probability that the other side is also black.
Step-by-step explanation:
Here let B1: Event of picking a card that has a black side
B2: Event of picking a card that has BOTH black side.
Now, by the CONDITIONAL PROBABILITY:
Now, as EXACTLY ONE CARD has both sides BLACK in three cards.
⇒ P (B1 ∩ B2) = 1 /3
Also, Out if total 6 sides of cards, 3 are BLACK from one side.
⇒ P (B1 ) = 3 /6 = 1/2
Putting these values in the formula, we get:
⇒ P (B2 / B1) = 2/3
Hence, there is a 2/3 probability that the other side is also black.
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Answer:
(a) -3/4
(b) -0.75
(c) -0.75
Step-by-step explanation:
It's a bit hard to tell what constitutes an "iteration" when using the bisection method to approximate a polynomial root. For the purpose here, we'll say one iteration consists of ...
- evaluating the function at the midpoint of the bracketing interval
- choosing a smaller bracketing interval
- identifying the x-value known to be closest to the solution
Thus, the result of the iteration consists of a bracketing interval and the choice of one of the interval's ends as the solution approximation.
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(a) We observe that the graphs intersect in the interval (-1, 0). For the first iteration, we evaluate f(x)-g(x) at x=-1/2. This tells us the solution is in the interval (-1, -1/2). The x-value closest to the root is x=-1/2.
For the second iteration, we evaluate the function f(x)-g(x) at x=-3/4. This tells us the solution is in the interval (-1, -3/4). The x-value closest to the root is x=-3/4.
For the third iteration, we evaluate the function f(x)-g(x) at x=-7/8. This tells us the solution is in the interval (-7/8, -3/4). The x-value closest to the root is x=-3/4.
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(b) The graph tells us the solution is approximately 0.7549. Rounded to 2 decimal places, the solution is approximately 0.75.
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(c) The above solution found after 3 iterations rounded to 2 decimal places is exactly 0.75.
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See the attached table for function values.
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<em>Comment on bisection iteration</em>
Since you cut the interval containing the root in half with each iteration, you gain approximately one decimal place for each 3 iterations. When the function value is very nearly zero at one of the interval endpoints, it can take many more iterations to achieve a better result.
Here, it takes 4 more iterations before an x-value becomes closer to the solution (x≈-97/128). And it takes one more iteration to move the end of the interval away from -3/4. After these 5 more iterations (8 total), the solution is known to lie in the interval (-97/128, -193/256). The corresponding solution approximation is -193/256. It is still only correct to 2 decimal places.
