This Question: 1 pt 3 of 13 (2 complete) This Test: 13 pts possibl In the game o

Question

This Question: 1 pt 3 of 13 (2 complete) This Test: 13 pts possibl In the game of roulette, a player can place a S6 bet on thenmber 27 and have a 38 probability of winning. If the metal ball lands on 27, the player gets to keep the $6 paid to play the game and the player is awarded $210. Otherwise, the player is awarded nothing and the casino takes the players $6. What is the expected value of the game to the player? If you played the game 1000 times, how much would you expect to lose? The expected value is Round to the nearest cent as needed.) The player would expect to lose about s (Round to the nearest cent as needed.)

Explanation / Answer

P(winning) = 1/38

P(losing) = 1-1/38 = 37/38

E(x) = 210*(1/38) + (-6)*(37/38)

= -12/38 = -0.3158

The expected value is -0.3158

The player would expect to lose about = 1000*(-0.3158) = 315.80

2)

As per binomial distribution,

P(X=r) = nCr * p^r * (1-p)^(n-r)

Here, n=6 , p =0.7

P(X =0) = 6C0 *0.7^0 * 0.3^6 = 0.0007

P(X =1) = 6C1 *0.7^1 * 0.3^5 = 0.0102

P(X =2) = 6C0 *0.7^2 * 0.3^4 = 0.0595

P(X =3) = 6C0 *0.7^3 * 0.3^3 = 0.1852

P(X =4) = 6C0 *0.7^4 * 0.3^2 = 0.3241

P(X =5) = 6C0 *0.7^5 * 0.3^1 = 0.3025

P(X =6) = 6C0 *0.7^6 * 0.3^0 = 0.1176

x P(x)

0 0.0007

1 0.0102

2 0.0595

3 0.1852

4 0.3241

5 0.3025

6 0.1176

b)

mux =0*0.0007+1*0.0102+2*0.0595+3*0.1852+4*0.3241+5*0.3025+6*0.1176

= 4.1993

mux = 4.19

sigmax = (0-4.1993)^2*0.0007+(1-4.1993)^2*0.0102+(2-4.199)^2*0.0595+(3-4.1993)^2*0.1852+(4-4.1993)^2*0.3241+(5-4.1993)^2*0.3025+(6-4.1993)^2*0.1176

sigmax = 1.26

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