Final answer:
To calculate the conditional probability that Harold made it to the cruise given that only one made it, we apply the concept of independent events. The probability of Harold making it is found to be approximately 51.21%.
Step-by-step explanation:
The subject of this question is probability, which falls under the category of Mathematics. To find the probability that Harold was the one who made it given that only one person made it to the cruise, we'll use the concept of conditional probability. We are given that Harold's chance of making it is 87% (0.87) and Maude's is 91% (0.91), and the events are independent.
Firstly, we calculate the probability of the event "only one of them makes it to the cruise":
- Probability that only Harold makes it (H and not M): 0.87 * (1 - 0.91)
- Probability that only Maude makes it (not H and M): (1 - 0.87) * 0.91
Then sum these two probabilities to get the probability of the event "only one makes it":
Probability of only one making it = P(H and not M) + P(not H and M)
Now, we apply the formula for conditional probability, where we want the probability that Harold was the one who made it given that only one made it (P(Harold|one made it)):
P(Harold|one made it) = P(Harold and not Maude) / P(only one made it)
Plugging in our calculated values:
P(Harold|one made it) = [0.87 * (1 - 0.91)] / [0.87 * (1 - 0.91) + (1 - 0.87) * 0.91]
By performing the calculations:
P(Harold|one made it) = 0.087 / (0.087 + 0.0829)
P(Harold|one made it) = 0.087 / 0.1699
P(Harold|one made it) ≈ 0.5121
Therefore, there is approximately a 51.21% chance that Harold was the one who made it to the cruise given that only one of them made it.