Suppose Mr. Smith goes to Vegas. He knows from previous experiences that the probability he will come out ahead after playing black jack is .3. The probability he will be ahead after playing the roulette wheel is .4. He also knows that his chances of coming out ahead in one game is independent of the other. If he plays both games, find the probability he will end up being ahead in at least one.
Hi late4thing,
Always remember that P(atleast one) = 1 - P(none)
Now the probability of Mr. Smith being ahead in blackjack is P(B) = 0.3. The probability of Mr. Smith not being ahead in blackjack will be P(not B) = 0.7.
The probability of Mr. Smith being ahead in roulette is P(R) = 0.4. The probability of Mr. Smith not being ahead in roulette will be P(not R) = 0.6.
The questions asks us to find the probability of Mr. Smith being ahead in at least one out of blackjack and roulette.
P(being ahead in atleast one) = 1 - P(being ahead in none)
= 1 - (0.7 * 0.6)
= 1 - 0.42
= 0.58
Hope this helps!
CrackVerbal Academics Team
