\$Let, y = 1/x\$
\$dy/dx = d/dx(1/x) = d/dx(x^-1) = x^(-1-1) = x^-2 = 1/x^2\$
\$dy/dx = d/dx(1/x) = d/dx(x^-1) = x^(-1-1) = x^-2 = 1/x^2\$
Derivative of 1/x & Tossing a coin problem
Upasana | August 31, 2019 | 2 min read | 103 views
What is the derivative of 1/x?
For finding derivative of 1/x, there is a condition if x is real number. In case x is real & x > 0 then,
Tossing a coin ten times resulted in 8 heads and 2 tails. How would you analyze whether a coin is fair? What is the p-value?
Answer :
Analyzing is coin is fair by tossing coin only 10 times would not be a fair decision as precision would be two low. As if we go by hypothesis then we must reject the hypothesis that coin is fair if p-value is too low.
Here, p-value is low so coin is not fair.
You have 10 coins. You toss each coin 10 times (100 tosses in total) and observe results. Would you modify your approach to the the way you test the fairness of coins?
Since, we will be tossing coin atleast 100 times, we shall see if it obeys null hypothesis or not. Here null hypothesis will be coin is unbiased means probability of getting heads should be qual to probability of getting tails.
There are steps that we can follow to find if a coin is fair or biased. This is all based on probability so you may have guessed the process already.
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Toss the coin 100 times.
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If total number of heads is not equal to total number of tails in those X times, then coin could be biased since probability of getting head is not equal to probability of getting tail.
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Though with X increasing, both probabilities shall approach closer to each other if coin is fair.
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