(10) If a random variable assumes 4 values with probability 1+3x 1-x 1+2xand4the condition on x so that these values represent the probability function of X is:(a) x =314(b) x =OII1|21-4x4(d) x = 0

Here for given probability check whether using value of x total probability is equal to 1 and all…

Answered: 1-4x 10) If a random variable assumes 4…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of tails in each outcome. For example, if the outcome is hth, then N (hth) = 1. Suppose that the random variable X is defined in terms of N as follows: X=2N² -6N-1. The values of Xare given in the table below. Outcome thh tth hhh hth ttt htt hht tht Value of X -5 -5 − 1 -5 −1 -5 -5 -5 Calculate the probabilities P(X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value X of X P(X=x) 0 8 X
I roll a fair die twice and obtain two numbers: X1 = result of the first roll, and X2 = result of the second roll. Write down the sample space S, and assuming that all outcomes are equally likely (because the die is fair), find the probability of the event A defined as the event that X1 + X2 = 8.
An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is thh, then N (thh)=2. Suppose that the random variable X is defined in terms of N as follows: X=2N²-6N-4. The values of X are given in the table below. ttt hhh hth hht tht htt thh tth Value of X -4 -4 -8-8-8-8-8-8 Outcome Calculate the probabilities P (X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value x of X P(X=x) 7 00 X S
An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is hhh, then N (hhh) = = 3. Suppose that the random variable X is defined in terms of N as follows: X=6N-2N²-3. The values of X are given in the table below. Outcome hhh hth hht thh htt tth ttt tht Value of X-3 1 1 1 1 1 -3 1 Calculate the probabilities P (X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value X of X P(X=x) 0 0 0 00 X
An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (*) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is ttt, then N (ttt) = 0. Suppose that the random variable X is defined in terms of N as follows: X=2N -2. The values of X are given in the table below. Outcome ttt hth tht htt thh hhh hht tth Value of X -2 2 0 0 2 4 2 0 Calculate the probabilities P(X=*) of the probability distribution of X. First, fill in the first row with the valuesof X. Then fill in the appropriate probabilities in the second row. Value x of X ___ ___ ___ ___ P(x=x) ___ ___ ___ ___
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An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of tails in each outcome. For example, if the outcome is hth, then N (hth) = 1. Suppose that the random variable X is defined in terms of N as follows: X=2N² − 6N-1. The values of X are given in the table below. Outcome thh tth hhh hth ttt htt hht tht Value of X-5 -5 -1 -5 -1 -5 -5 -5 Calculate the probabilities P(X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value X of X P(X=x) 0 0 00 X
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1-4x 10) If a random variable assumes 4 values with probability 1+3x 1-x 1+2x and ¹-4 4 4 4 the condition on x so that these values represent the probability function of X is: