A player tosses a die 6 times. If gets a number 6 Atleast two times he wins 2 dollars ,otherwise he looses 1 dollar.. Find the expectation of the amount if he wins

Step-by-step explanation:

The expected value for a discrete variable is calculated as:

Where and are the values that the variable can take and and are their respective probabilities.

So, a player can win 2 dollars or looses 1 dollar, it means that is equal to 2 and is equal to -1.

Then, we need to calculated the probability that the player win 2 dollars and the probability that the player loses 1 dollar.

If there are n identical and independent events with a probability p of success and a probability (1-p) of fail, the probability that a events from the n are success are equal to:

Where

So, in this case, n is number of times that the player tosses a die and p is the probability to get a 6. n is equal to 6 and p is equal to 1/6.

Therefore, the probability   that a player get at least two times number 6, is calculated as:

On the other hand, the probability that a player don't get  at least two times number 6, is calculated as:

Finally, the expected value of the amount that the player wins is:

It means that he can expect to loses 0.2101 dollars.

it is b)

hope this : )

step-by-step explanation:

i got 10.28 but that rounds up to 10.3