A fair coin is tossed three times. A player wins $1 if the first toss is a head, but loses $1 if the first toss is a tail. Similarly, the player wins $2 if the second toss is a head, but loses $2 if the second toss is a tail, and wins or loses $3 according to the result of the third toss. Let the random variable X be the total winnings after the three tosses (possibly a negative value if losses are incurred). (a) Construct the probability mass function. (b) Construct the cumulative distribution function. (c) What is the most likely value of the random variable X?

Step-by-step explanation:

First lets construct the outcome tree based on the winning situation in the question;

the coin was toasted 3 times

so

FIRST                      SECOND                THIRD                         TOTAL

toss  winning         toss  winning          toss  winning             winning

H         1                   H           2                 H         3                        6

H         1                   H           2                 T        -3                        0

H         1                   T          -2                  H        3                        2

H         1                   T          -2                  T       -3                       -4        

T        -1                    H          2                  H        3                        4

T        -1                    T          2                   H      -3                       -2

T        -1                    T         -2                   H       3                        0

T        -1                    T         -2                   H      -3                       -6

NOW

a) the probability mass function is;

Xi     -6       -4       -2        0        2        4       6

Pi     1/8     1/8      1/8      2/8     1/8     1/8     1/8

b) the cumulative distribution function

Xi        -6       -4       -2        0        2        4       6

Pi        1/8     1/8      1/8      2/8     1/8     1/8     1/8

F(Xi)    1/8    2/8     3/8      5/8     6/8    7/8    8/8    

c) most likely value of the random variable X

The most likely value of the random variable X is 0

from the probability mass function in (a), the probability the random variable is highest at 2/8, te Xi value above 2/8 is 0

therefore the most likely value of the random variable X is 0

the correct answer is a.) bc< mn 5 stars if this you : )

step-by-step explanation: