1. {Expected value calculation, discrete random event} Suppose you have a coin t

Question

1. {Expected value calculation, discrete random event}

Suppose you have a coin that lands on Heads with probability 0.6. Where the random variable X is the sum of two coin flips, (where heads count 1 and tails count 0 - two heads then X= 2, a head and a tail then X = 1 and two tails the X = 0). Compute the expected value of X.

2. {Expected value calculation, discrete random event}

Suppose you have a fair coin. Where the random variable X is the sum of three coin flips, (where heads count 1 and tails count 0 - three heads then X= 3, two heads and a tail is X = 2, just one head has X=1, and with three tails X = 0). Compute the expected value of X.

Explanation / Answer

(1) RADOM VARIABLE `X` WIL TAKE VALUES 2, 1, 1 AND 0 AT SAMPLE POINT { HH, HT, TH, TT } WITH RESPECTIVE PROBABILITY 0.36, 0.24,0.24 AND 0.16 AS PROBABILITY OF COMING HEAD IS 0.6 AND TAIL IS 0.4 (GIVEN).

NOW X1=2, PROBABILITY P1= 0.36 X3=1, P3=0.24

X2=1 PROBABILITY P2= 0.24 X4=0, P4= 0.16 HENCE EXPECTED VALUE OF `X` WILL BE GIVEN AS E(X) = P1X1+P2X2+P3X3+P4X4

= 0.36 (2)+ 0.24(1) +0.24(1) +0.16(0)

= 0.72 + 0.24 + 0.24 + 0 = 1.22

(2) FOR A FAIR COIN PROBABILITY OF COMING HEAD IS `1/2` AND TAIL IS `1/2`

RANDOM VARIABLE`X`(SUM OF THREE COIN FLIPS) WILL TAKE VALUES 3,2,1 AND `0` WITH SAMPLE POINT { HHH, (HHT,HTH,THH), (THT,TTH,HTT), TTT } WITH RESPECTIVE PROBABILITY 1/8 ,3/8, 3/8 AND 1/8

NOW X1=3, P1=1/8 , X2=2, P2= 3/8 . X3= 1, P3= 3/8 AND X4= 0, P4= 1/8 NOW EXPECTED `X` IS GIVEN AS

E(X) = P1X1+P2X2+P3X3+P4X4 = 1/8 (3) + 3/8(2) +3/8 (1) + 1/8 (0)

= 3/8 +6/8 +3/8 +0 = 12/8 = 1.5

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