Ming rolls a number cube, tosses a coin and chooses a card from two cards marked A and B. If an even number and heads appear, Ming wins, no matter which card is chosen. Otherwise Lashonda wins. Find P (Ming wins)
P(wing wins)=3/6*1/2=1/4
To find the probability that Ming wins, we first need to determine the probability of two independent events happening: rolling an even number on the number cube and getting heads on the coin toss.
The number cube has six possible outcomes, as it has six sides numbered 1 to 6. Out of these six outcomes, half of them are even numbers (2, 4, and 6), and the other half are odd numbers (1, 3, and 5). So the probability of rolling an even number is 3 out of 6, or 1/2.
Similarly, the coin has two possible outcomes: heads or tails. Since we want to find the probability of getting heads, and assuming the coin is fair, there is a 1 out of 2 chance of getting heads, or 1/2.
Since these two events are happening independently, we can multiply the probabilities. So the probability of rolling an even number and getting heads is (1/2) * (1/2) = 1/4.
Now we need to consider that if an even number and heads appear, Ming wins no matter which card is chosen. This means that Lashonda will only win if an odd number or tails appear. Since the probability of Ming winning is the opposite of Lashonda winning, we can say that P(Ming wins) = 1 - P(Lashonda wins).
The probability of Lashonda winning is the complement of Ming winning, which means it is 1 - 1/4 = 3/4.
Therefore, P(Ming wins) = 1 - P(Lashonda wins) = 1 - 3/4 = 1/4.