joseph and his friends were planning their trip to a water park. the park offers 41 total rides, including…

joseph and his friends were planning their trip to a water park. the park offers 41 total rides, including 36 speed slides and 3 funnel water slides. if joseph randomly chooses to go on 37 of the rides, what is the probability that 35 of the chosen rides are speed slides and 2 are funnel water slides? write your answer as a decimal rounded to four decimal places.
Answer
Explanation:
Step1: Calculate total combinations
We use the combination formula $C(n,k)=\frac{n!}{k!(n - k)!}$, where $n$ is the total number of items and $k$ is the number of items to choose. The total number of ways to choose 37 rides out of 41 is $C(41,37)=\frac{41!}{37!(41 - 37)!}=\frac{41!}{37!4!}=\frac{41\times40\times39\times38}{4\times3\times2\times1}=91390$.
Step2: Calculate favorable combinations
The number of ways to choose 35 speed - slides out of 36 is $C(36,35)=\frac{36!}{35!(36 - 35)!}=36$. The number of ways to choose 2 funnel water - slides out of 3 is $C(3,2)=\frac{3!}{2!(3 - 2)!}=3$. The number of favorable combinations is the product of these two combinations, so $C(36,35)\times C(3,2)=36\times3 = 108$.
Step3: Calculate probability
The probability $P$ is the number of favorable combinations divided by the total number of combinations. So $P=\frac{108}{91390}\approx0.0012$.
Answer:
0.0012