(a)
To find:
The total number of possible outcomes when a coin is flipped twice.
(b)
To find:
The total number of possible outcomes when a coin is flipped three times.
(c)
To find:
The total number of possible outcomes when a coin is flipped four times.
(d)
To find:
The total number of possible outcomes when a coin is flipped then a die is rolled.
(e)
To find:
The total number of possible outcomes when two dice are rolled.
(f)
To find:
The total number of possible outcomes when two coins are flipped, then a die is rolled.
(g)
To find:
The total number of possible outcomes when a coin is flipped, then a card is drawn.
(h)
To find:
The total number of possible outcomes when a die is rolled, then a card is drawn.
(i)
To find:
The total number of possible outcomes when a coin is flipped, then a die is rolled and then a card is drawn.
(j)
To find:
The total number of possible outcomes when two cards are drawn with replacement.
(k)
To find:
The total number of possible outcomes when two cards are drawn without replacement.
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PROBLEM SOLVING APPROACH (PRINT UPGRADE)
- 4. Prove that, for any random variable X, the minimum of EIX-al is attained for a = med (X).arrow_forward8. Recall, from Sect. 2.16.4, the likelihood ratio statistic, Ln, which was defined as a product of independent, identically distributed random variables with mean 1 (under the so-called null hypothesis), and the, sometimes more convenient, log-likelihood, log L, which was a sum of independent, identically distributed random variables, which, however, do not have mean log 1 = 0. (a) Verify that the last claim is correct, by proving the more general statement, namely that, if Y is a non-negative random variable with finite mean, then E(log Y) log(EY). (b) Prove that, in fact, there is strict inequality: E(log Y) < log(EY), unless Y is degenerate. (c) Review the proof of Jensen's inequality, Theorem 5.1. Generalize with a glimpse on (b).arrow_forward2. Derive the component transformation equations for tensors shown be- low where [C] = [BA] is the direction cosine matrix from frame A to B. B[T] = [C]^[T][C]T 3. The transport theorem for vectors shows that the time derivative can be constructed from two parts: the first is an explicit frame-dependent change of the vector whereas the second is an active rotational change of the vector. The same holds true for tensors. Starting from the previous result, derive a version of transport theorem for tensors. [C] (^[T])[C] = dt d B dt B [T] + [WB/A]B[T] – TWB/A] (10 pt) (7pt)arrow_forward
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