PROBLEM 1 [30 POINTS = 5 + 5 + 5 + 5 + 5 + 5] TOM AND JERRY DECIDED TO PLAY A MA

Question

PROBLEM 1 [30 POINTS = 5 + 5 + 5 + 5 + 5 + 5] TOM AND JERRY DECIDED TO PLAY A MATCH OF 6 TENNIS GAMES. HISTORICALLY THEY HAVE ASSESSED JERRY’S WINNING CHANCES AS 40%, WHILE TOM WINS 60% OF THEIR GAMES. CONSIDER A RANDOM VARIABLE (W) THAT SHOWS THE NUMBER OF GAMES WON BY TOM. THEREFORE (6 – W) IS THE NUMBER OF GAMES LOST BY TOM. SEPARATE GAMES ARE VIEWED AS INDEPENDENT TRIALS. SHOW THE FORMULA THAT YOU USE IN EACH CASE. 1. WHAT IS THE CHANCE THAT TOM WINS AT MOST 5 OUR OF 6 GAMES? P [W 5] = 2. WHAT IS THE CHANCE THAT TOM WINS AT LEAST 1 OUT OF 6 GAMES? P [W 1] = 3. HOW MANY GAMES DOES TOM EXPECT TO WIN? E [W] = 4. WHAT IS THE VARIANCE OF THE RANDOM VARIABLE W? VAR [W] = 5. IF TOM IS GAINING $3 FOR EACH VICTORY AND LOSING $4 FOR EACH GAME HE HAS LOST, HOW MUCH MONEY DOES HE EXPECT TO WIN? FIND THE EXPECTATION OF Z = [3 W – 4 (6 – W)] = 7W – 24. E [Z] = ALSO DETERMINE THE VARIANCE AND STANDARD DEVIATION OF Z. VAR [Z] = SD [Z] = 6. WHAT IS THE STANDARD DEVIATION OF (6 – W), THE NUMBER OF GAMES LOST BY TOM? SD [6 – W] = HOMEWORK 5 – STAT 3360 – SPRING

Explanation / Answer

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Probability of Tom winning = 0.6

Probability of Jerry winining = 0.4

a) Tom wins at most five games

P(Tom wins at most five games) = 1 - P(tom wins all the games)

=> 1 - 6C6 * (0.6)^6 * (0.4)^0

=> 1 - (0.6)^6

=> 0.953344

b)

P(tom win atleast 1 out of 6 games) = 1 - P(tom wins no games)

=> 1 - 6C6 * (0.4)^6 * (0.6)^0

=> 1 - (0.4)^6

=> 0.995904

c)

The number of games top is expced to win is

E[X] = probability of winning the game * number of games

=> 0.6 * 6

=> 3.6

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