Sponsors of a local charity decided to attract wealthy patrons to its $500-a-plate dinner by allowing each patron to buy a set of 20 tickets for the gaming tables. The chance of winning a prize for each of the 20 plays is 50–50. If you bought 20 tickets, what is the chance of winning 15 or more prizes?

Answer:There is a 2.06% probability of winning 15 or more prizes.Step-by-step explanation:For each play, there are only two possible outcomes. Either you win a prize, or you do not win a prize. This means that we can solve this problem using concepts of the binomial probability distribution.Binomial probability distributionThe binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]And p is the probability of X happening.In this problem we have that:There are 20 plays, no [tex]n = 20[/tex]The chance of winning a prize for each of the 20 plays is 50–50, so [tex]p = 0.50[/tex].If you bought 20 tickets, what is the chance of winning 15 or more prizes?This is [tex]P(X \geq 15)[/tex][tex]P(X \geq 15) = P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)[/tex][tex]P(X = 15) = C_{20,15}.(0.50)^{15}.(0.50)^{5} = 0.0148[/tex][tex]P(X = 16) = C_{20,16}.(0.50)^{16}.(0.50)^{4} = 0.0046[/tex][tex]P(X = 17) = C_{20,17}.(0.50)^{17}.(0.50)^{3} = 0.001[/tex][tex]P(X = 18) = C_{20,18}.(0.50)^{18}.(0.50)^{2} = 0.0002[/tex][tex]P(X = 19) = C_{20,19}.(0.50)^{19}.(0.50)^{1} = 0.00002[/tex][tex]P(X = 20) = C_{20,20}.(0.50)^{20}.(0.50)^{0} = 0.00000009[/tex]So[tex]P(X \geq 15) = P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0148 + 0.0046 + 0.001 + 0.0002 + 0.00002 + 0.00000009 = 0.0206[/tex]There is a 2.06% probability of winning 15 or more prizes.