Number and Algebra
Functions
Geometry and Trigonometry
Statistics and Probability
Calculus
The random variable XXX takes the value 222 with probability ppp and −3-3−3 with probability 1−p1-p1−p. Find ppp such that the expected value is zero.
The random variable XXX takes values 111 with probability ppp and −1-1−1 with probability 1−p1-p1−p. Find ppp so that the game is fair (i.e. E[X]=0E[X]=0E[X]=0).
The random variable XXX has the distribution P(X=0)=p,P(X=1)=2p,P(X=2)=1−3p.\quad P(X=0)=p, \\ \quad P(X=1)=2p,\\ \quad P(X=2)=1-3p. P(X=0)=p,P(X=1)=2p,P(X=2)=1−3p. Find ppp so that this is a valid probability distribution.
Let XXX have P(X=−1)=p,P(X=0)=2p,P(X=2)=1−3p.\quad P(X=-1)=p, \\ \quad P(X=0)=2p, \\ \quad P(X=2)=1-3p. P(X=−1)=p,P(X=0)=2p,P(X=2)=1−3p. Find ppp such that PPP is a valid distribution.
The random variable XXX takes values −2-2−2, 111, 333 with probabilities ppp, qqq, and 1−p−q1-p-q1−p−q, respectively. Find ppp and qqq so that the game is fair (E[X]=0E[X]=0E[X]=0) and the probabilities sum to 1.
The random variable XXX takes values −2,−1,0,1,2-2,-1,0,1,2−2,−1,0,1,2 with probabilities P(X=−2)=p,P(X=−1)=2p,P(X=0)=1−6p,P(X=1)=2p,P(X=2)=p.P(X=-2)=p, \quad P(X=-1)=2p, \quad P(X=0)=1-6p, \quad P(X=1)=2p, \quad P(X=2)=p. P(X=−2)=p,P(X=−1)=2p,P(X=0)=1−6p,P(X=1)=2p,P(X=2)=p. Find the range of ppp for which this is a valid probability distribution.
The random variable XXX takes values 000, 333, and 555 with probabilities ppp, qqq, and 1−p−q1-p-q1−p−q, respectively. Find ppp and qqq such that E[X]=4E[X]=4E[X]=4 and the probabilities sum to 1.
Let XXX have P(X=−2)=p,P(X=0)=2q,P(X=5)=1−p−2q.P(X=-2)=p, \quad P(X=0)=2q, \quad P(X=5)=1-p-2q. P(X=−2)=p,P(X=0)=2q,P(X=5)=1−p−2q. Find ppp and qqq so that E[X]=1E[X]=1E[X]=1 and the probabilities sum to 1.
The variable XXX takes values 1,2,3,41,2,3,41,2,3,4 with probabilities P(X=1)=p,P(X=2)=2p,P(X=3)=3p,P(X=4)=1−6p.P(X=1)=p, \quad P(X=2)=2p, \quad P(X=3)=3p, \quad P(X=4)=1-6p. P(X=1)=p,P(X=2)=2p,P(X=3)=3p,P(X=4)=1−6p. Find ppp such that E[X]=3E[X]=3E[X]=3 and the distribution is valid.
The distribution of XXX is P(X=−1)=2p,P(X=2)=3q,P(X=4)=1−2p−3q.P(X=-1)=2p, \quad P(X=2)=3q, \quad P(X=4)=1-2p-3q. P(X=−1)=2p,P(X=2)=3q,P(X=4)=1−2p−3q.Find ppp and qqq such that the game is fair (E[X]=0E[X]=0E[X]=0) and probabilities sum to 1.
XXX takes four values −2,0,1,3-2,0,1,3−2,0,1,3 with probabilities P(X=−2)=p,P(X=0)=q,P(X=1)=2p,P(X=3)=1−p−q−2p.\quad P(X=-2)=p, \\ \quad P(X=0)=q, \\ \quad P(X=1)=2p, \\ \quad P(X=3)=1-p-q-2p. P(X=−2)=p,P(X=0)=q,P(X=1)=2p,P(X=3)=1−p−q−2p. Find p,qp,qp,q so that the game is fair (E[X]=0E[X]=0E[X]=0) and probabilities sum to 1.
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