chapter 1

Chapter 3: Discrete Random Variables and Probability Distributions21. The jumps in F(x) occur at x = 0, 1, 2, 3, 4, 5, and 6, so we first calculate F( ) at each of thesevalues:F(0) = P(X ≤ 0) = P(X = 0) = .10F(1) = P(X ≤ 1) = p(0) + p(1) = .25F(2) = P(X ≤ 2) = p(0) + p(1) + p(2) = .45F(3) = .70, F(4) = .90, F(5) = .96, and F(6) = 1.The c.d.f. isF(x) =⎧ .00⎪⎪.10⎪ .25⎪.45⎨⎪ .70⎪ .90⎪⎪ .96⎪⎩1.00x < 00 ≤ x < 11 ≤ x < 22 ≤ x < 33 ≤ x < 44 ≤ x < 55 ≤ x < 66 ≤ xThen P(X ≤ 3) = F(3) = .70, P(X < 3) = P(X ≤ 2) = F(2) = .45,P(3 ≤ X) = 1 – P(X ≤ 2) = 1 – F(2) = 1 - .45 = .55,and P(2 ≤ X ≤ 5) = F(5) – F(1) = .96 - .25 = .7122.a. P(X = 2) = .39 - .19 = .20b. P(X > 3) = 1 - .67 = .33c. P(2 ≤ X ≤ 5) = .92 - .19 = .78d. P(2 < X < 5) = .92 - .39 = .5323.a. Possible X values are those values at which F(x) jumps, and the probability of anyparticular value is the size of the jump at that value. Thus we have:x 1 3 4 6 12p(x) .30 .10 .05 .15 .40b. P(3 ≤ X ≤ 6) = F(6) – F(3-) = .60 - .30 = .30P(4 ≤ X) = 1 – P(X < 4) = 1 – F(4-) = 1 - .40 = .6024. P(0) = P(Y = 0) = P(B first) = pP(1) = P(Y = 1) = P(G first, then B) = P(GB) = (1 – p)pP(2) = P(Y = 2) = P(GGB) = (1 – p) 2 pContinuing, p(y) = P(Y=y) = P(y G’s and then a B) = (1 – p) y p for y = 0,1,2,3,…101