Question

Solve the problem. In a game of roulette, Jorge places 170 bets of Sl each on the number 3. A win pays off with odds 35 1 and on any one spin there is a 1/38 probability that 3 will be the winning number. Among the 170 bets, what is the minimum number of wins needed for Jorge to make a profit? Estimate the probability that Jorge will make a profit. A 5:0.4013 3.6:0.3121 C.5;0.496 0.6; 0.2327

Answer 1

Answer:

Total cost = 170x1 = $170

payoff is $35 per $1 bet

Number of wins needed to make a profit = 170/35

= 4.86 $\approx$ 5

P(win), p = 1/38

n = 170

P(Jorge will make a profit) = P(at least 5 wins)

mean = np = 4.47

standard deviation = $\sqrt{npq}$ = 2.09

P(X $\geq$ 5) = 1 - P(X < 5)

P(X < A) = 1 - P(Z < (A - mean)/standard deviation)

After applying continuity correction, P(X $\geq$ 5) = 1 - P(Z < (4.5 - 4.47)/2.09)

= 1 - P(Z < 0.01)

= 1 - 0.5040

= 0.496

So, the answer is C) 5; 0.496

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