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 Dec8 awarded Caucus Sep15 comment Binomial Theorem Application Exercise Hint: $\binom{n}{k}2^{n-k}=\binom{n}{k}\left(1+1\right)^{n-k}=\sum_{j=0}^{n-k}\binom{n‌​}{k}\binom{n-k}{j}$. Jul8 comment Probability Question involving Probability Mass Function/Random Variables You've already determined that $P(K=k)=\binom{n}{k}p^k(1-p)^{n-k}$. Part (b) is asking you to solve for $n$ such that $P(K\geq1)\geq0.95$. Another way to look at this is that the probability of not receiving the message correctly at all should be less than 5%: $P(K=0)=(1-p)^{n}\leq0.05$. May28 answered If $P(x) = ax^2 + bx + c$ and $Q(x) = -ax^2 + dx + c$, then prove that $P(x) \cdot Q(x) = 0$ has at least two real roots? Mar25 awarded Critic Nov16 comment Lottery probability question… Second place needs to capture exactly five of the six numbers. How many ways can this be done? The denominator remains the same: there are still $\binom{59}{6}$ ways to draw six numbers from 59. However, the numerator should count the number of ways you may draw five numbers from the winning six and one number from the remaining 53 numbers. May15 awarded Caucus Feb24 comment Given $x(t) = u(t)$ and $h(t) = \cos(\pi t)u(t)$, how do we find the response $y(t)$? I don't see why the integral isn't $\int_{0}^{\infty}cos(\pi\tau)u(\tau)u(t-\tau)d\tau$. Nov7 answered Poisson Distribution for Consecutive Figures Nov2 comment Solve Ax = b where b are labels instead of values @locke14 I misunderstood the question. My response was for $A$ having these entries, not $b$. Nov2 comment Solve Ax = b where b are labels instead of values Usually with categorical variables, one category is chosen as a baseline and all others are assigned their own indicator variable in the regression. For example, if the original variable represents weight and the categories are "low", "medium" and "high", we may choose "medium" as the baseline and create two other variables named "isLow" and "isHigh". If a person has "medium" weight, isLow and isHigh will both be zero. If a person has "low" weight, isLow will be one and isHigh will be zero. If a person has "high" weight, isLow will be zero and isHigh will be one. Oct30 comment Process for $(k+1)^3$? @wj32 Ah. That makes sense. Oct30 comment Process for $(k+1)^3$? You don't mention the binomial theorem anywhere in your question. Oct30 comment Process for $(k+1)^3$? Look into the binomial theorem. Oct28 comment Puzzle on Ranks The computation is correct. This is a good way of thinking about the problem though. Oct26 accepted Help Needed: Partial Derivative Identity/Chain Rule Oct26 comment Formulas for the multivariate Gaussian function? If you convert the density function to use the actual elements of $x$ and $\Sigma$ rather than the matrix equivalents, taking derivatives will become more clear. Oct17 comment Calculating Variance of a binomial distribution using the standard formula $E(X^2) - \mu^2$ Yes, that is correct. Oct17 revised Calculating Variance of a binomial distribution using the standard formula $E(X^2) - \mu^2$ added 186 characters in body Oct17 answered Calculating Variance of a binomial distribution using the standard formula $E(X^2) - \mu^2$