SOLUTION: A biased six-sided die has one face bearing number 3, two faces number 2 and three faces number 1. A boy tosses the die once and observes the number on the uppermost face. He then

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: A biased six-sided die has one face bearing number 3, two faces number 2 and three faces number 1. A boy tosses the die once and observes the number on the uppermost face. He then       Log On

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 Click here to see ALL problems on Probability-and-statistics Question 630570: A biased six-sided die has one face bearing number 3, two faces number 2 and three faces number 1. A boy tosses the die once and observes the number on the uppermost face. He then tosses a number of fair coins equal to the number shown on the die. The random variable X is the number of heads obtained. Construct the probability density function of X.Answer by Edwin McCravy(8893)   (Show Source): You can put this solution on YOUR website!A biased six-sided die has one face bearing number 3, two faces number 2 and three faces number 1. A boy tosses the die once and observes the number on the uppermost face. He then tosses a number of fair coins equal to the number shown on the die. The random variable X is the number of heads obtained. Construct the probability density function of X. ``` I. P(1 on die) = 3/6 = 1/2 A. Possible tosses {H,T} B. P(1 toss 0 heads) = 1/2 C. P(1 toss 1 head) = 1/2 II. P(2 on die) = 2/6 = 1/3 A. Possible tosses {HH,HT,TH,TT} B. P(2 tosses 0 heads) = 1/4 C. P(2 tosses 1 head) = 2/4 = 1/2 D. P(2 tosses 2 heads) = 1/4 III. P(3 on die) = 1/6 A. Possible tosses {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} B. P(3 tosses 0 heads) = 1/8 C. P(3 tosses 1 head) = 3/8 D. P(3 tosses 2 heads) = 3/8 E. P(3 tosses 3 heads) = 1/8 P(x=0) = P(D1&H0)+P(D2&H0)+P(D3&H0) = (1/2)(1/2)+(1/3)(1/4)+(1/6)(1/8) = 17/48 P(x=1) = P(D1&H1)+P(D2&H1)+P(D3&H1) = (1/2)(1/2)+(1/3)(1/2)+(1/6)(3/8) = 23/48 P(x=2) = P(D2&H2)+P(D3&H2) = (1/3)(1/4)+(1/6)(3/8) = 7/48 P(x=3) = P(D3&H3) = (1/6)(1/8) = 1/48 Probability density function: X P(X) 0 17/48 1 23/48 2 7/48 3 1/48 Note that the probabilities have sum 48/48 or 1 Edwin```