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1 answer:
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Answer:
-$ 10.13
Step-by-step explanation:
Given,
The cost price of each ticket = $ 22,
So, the value of each ticket other than price ticket = - $ 22, ( negative sign shows loss,i.e. if we don't get the price we will have the loss of $ 22 )
Now, there are 3 tickets in which first is of $35,000 price, second is of $1,100 and third is of $ 600,
So, the value of first ticket = 35000 - 22 = $ 34978,
Value of second ticket = 1100 - 22 = $ 1078,
Value of third ticket = 600 - 22 = $ 578,
Also,
Hence, by the above information we can make a table for the given situation,
Number 2997 1 1 1
Price -$ 22 $ 34978 $ 1078 $ 578
Probability 2997/3000 1/3000 1/3000 1/3000
Therefore, the expected value of a ticket
≈ - $ 10.13