We expect to lose $0.37 per lottery ticket
<u>Explanation:</u>
six winning numbers from = { 1, 2, 3, ....., 50}
So, the probability of winning:


The probability of losing would be:
P(loss) = 1 - P(win)

According to the question,
When we win, then we gain $10 million and lose the cost of the lottery ticket.
So,
$10,000,000 - 1 = $9,999,999
When we lose, then we lose the cost of the lottery ticket = $1
The expected value is the sum of the product of each possibility x with its probability P(x):
E(x) = ∑ xP(x)

Thus, we expect to lose $0.37 per lottery ticket