To find the probability of at least 3 students carrying a mobile device when 4 are selected randomly, we can determine the probabilities of exactly 3 of them carrying one and all 4 carrying one, and then add those probabilities together.
Let S (students carrying mobile device) = 450 and T (total students) = 700
1. Probability that exactly 3 students have a mobile device and 1 student does not:
- Ways to pick 3 students with a mobile device (combinatorial formula: nCr): 450C3
- Ways to pick 1 student without a mobile device (combinatorial formula, removing the 450 mobile students from the total count): 250C1
- Ways to pick any 4 students from the total: 700C4
So, the probability for this case: (450C3 * 250C1) / 700C4
2. Probability that all 4 students have a mobile device:
- Ways to pick 4 students with a mobile device: 450C4
- Ways to pick any 4 students from the total: 700C4
So, the probability for this case: 450C4 / 700C4
Now, add the probabilities for both cases:
(450C3 * 250C1) / 700C4 + 450C4 / 700C4
Using the combinatorial formula, we get:
(450! / (3! * 447!)) * (250! / (1! * 249!)) / (700! / (4! * 696!)) + (450! / (4! * 446!)) / (700! / (4! * 696!))
Simplifying the expression, we get:
(11,198,100 * 250 + 67,425) / 84,270,245
Computing the numerical values, we get:
(2,799,525,000 + 67,425) / 84,270,245
2,799,592,425 / 84,270,245 = approximately 0.332308
So the probability that at least 3 out of 4 randomly selected students carry a mobile device is approximately 0.332308 or 33.23%.
If there are 700 students in a school and 450 of them carry a mobile device, what is the
probability that when 4 students are randomly selected, that at least 3 of them carry a mobile
device?
1 answer