In playing a game with an unfair coin, you win $5.00 with probability 1/3 but lose $2.00 with probability 2/3. Determine the number of games that you should play to at least break even with probability 0.99. questions

the number of games that you should play to at least break even with probability 0.99 is 30

Step-by-step explanation:

Given the data in the question;

we represent the total number of games by n

and among the n number of games, k is the number of wins.

so

P( at least  break even ) = P( k × $5 ≥ $2( n -k ) )

P( k × $5 ≥ $2( n -k ) ) = P( 5K ≥ 2n - 2k ) = P( 7k ≥ 2n )

so

P( at least  break even ) = P( 7k ≥ 2n ) ≥ 0.99

⇒ P( p' ≥ (2/7) ) ≥ 0.99

p' ~ Normal( Mean = 0.5, standard deviation = √( (0.5×0.5) / n )

so

P( Z ≥   ((2/7) - 0.5) / √( 0.25/n ) ≥ 0.99

⇒ ((2/7) - 0.5) / √( 0.25/n ) ≥ -2.326

⇒ (0.2857-0.5) / (0.5/√n)  ≥ -2.326

⇒ -0.4286 × √n ≥ -2.326

√n ≥ -2.326 / -0.4286

√n ≥ 5.42697

n ≥ (5.42697)²

n ≥ 29.35 since its sample size

It approximately becomes 30

Therefore, the number of games that you should play to at least break even with probability 0.99 is 30

mk./km: lm

step-by-step explanation:

,mlmlm

; mk

lk

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58