Suppose that in the game of Blackjack, a player has a 49% chance of winning any

Question

Suppose that in the game of Blackjack, a player has a 49% chance of winning any given hand. Furthermore, suppose a player is going to play 200 hands of Blackjack. If a player bets $50 on each hand (if they win the hand then they win $50 and if they lose the hand then they lose $50) then if the player wins fewer than 90 times (out of 200) then they lose more than $1000. What is the probability that over the next 200 hands of Blackjack that a loses more than $1000 (or what is the probability that a player wins fewer than 90 times of of the next 200 hands or 45% of the time)? (please round your answer to 4 decimal places)

Explanation / Answer

X be a random variable representing number of hands won

X follows binomial distribution

n = 200

p = 0.49

q = 0.51

P(X<90) = binomialcdf(89,200,0.49)

= 0.1145

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