Probability: Nando has a gambling problem, and plays the following game of chance:

iwannapassstats

New member
Joined
Nov 6, 2017
Messages
2
hey all, this question is really hard, need help ASAP

cheers :)



Nando has a gambling problem, and plays a game of chance:

She pays $1 to roll two six-sided dice (each with sides labelled 1 through 6). If she rolls a double (that is, if both dice display the same number), she wins $5. If she doesn't roll a double on her first roll, then she gets a second roll and wins $3 if she rolls a double on this second throw. If she doesn't roll a double on either throw, she wins nothing.

Let X be the random variable for the amount Nando gains from each game.

Paying $1 is important, as it changes the outcomes from $0, $3, and $5 to -$1, $2, and $4!

13. Find Nando's expected profit or loss if she plays 50 games.

(A) -$6.25. . . .(B) -$0.25. . . .(C) $0.25. . . .(D) $12.50. . . .(E) $62.50

14. Find the variance of her profit/loss if she plays ONE game [i.e., find Var(X)].

(A) 3.42. . . . ..(B) 3.85. . . . ..(C) 3.92. . . . .(D) 5.35. . . . . .(E) 5.42
 

Attachments

  • sjdjsijd.jpg
    sjdjsijd.jpg
    28.9 KB · Views: 16
Last edited by a moderator:
hey all, this question is really hard, need help ASAP

cheers :)



Nando has a gambling problem, and plays a game of chance:

She pays $1 to roll two six-sided dice (each with sides labelled 1 through 6). If she rolls a double (that is, if both dice display the same number), she wins $5. If she doesn't roll a double on her first roll, then she gets a second roll and wins $3 if she rolls a double on this second throw. If she doesn't roll a double on either throw, she wins nothing.

Let X be the random variable for the amount Nando gains from each game.

Paying $1 is important, as it changes the outcomes from $0, $3, and $5 to -$1, $2, and $4!

13. Find Nando's expected profit or loss if she plays 50 games.

(A) -$6.25. . . .(B) -$0.25. . . .(C) $0.25. . . .(D) $12.50. . . .(E) $62.50

14. Find the variance of her profit/loss if she plays ONE game [i.e., find Var(X)].

(A) 3.42. . . . ..(B) 3.85. . . . ..(C) 3.92. . . . .(D) 5.35. . . . . .(E) 5.42
Why the hurry?

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33
 
Last edited by a moderator:
so i know that the expected value also refers to the mean and so i thought of the equation
,W = aX + bY + c ---> Mean: uW = auX + buY + c but the values -1,2,4 dont fit in the corresponding coefficients
 
You're completely off in left field and WAY over thinking it. The problem has only one random variable, so why on Earth would the solution involve a linear combination of two random variables? Instead, let's return to the definitions of the terms. The expected value ("mean") of a random variable is:

\(\displaystyle E(X) = \displaystyle \sum_{x=1}^{n} \left( x * P(x) \right)\)

In other words, it is defined as the sum of the probabilities of each outcome, multiplied respectively by the value or "weight" associated with each outcome. Here, your random variable X has three possible outcomes:

  • Nando rolls doubles on her first toss
  • She rolls doubles on her second toss (Remember that this can only happen if she doesn't get doubles on her first toss)
  • She rolls doubles on neither toss

In the first outcome, Nando makes a profit of $4; in the second, she makes a profit of $2; in the third, she loses $1. So, find the probabilities of each outcome, plug them in, and crank away.

\(\displaystyle E(X) = 4 \cdot \text{ ??? } + 2 \cdot \text{ ??? } - 1 \cdot \text{ ??? }\)
 
Top