Probability Analysis of Student Performance Based on Gender
In a group of 600 students, there are 400 females and 200 males. Among these, 100 students have failed the exam, and 40 of them are male. Below is a tabular representation of the data:
| Gender | Failed (x) | Not Failed (x’) | Total |
|---|---|---|---|
| Female (F) | 40 | 360 | 400 |
| Male (F’) | 60 | 140 | 200 |
| Total | 100 | 500 | 600 |
Probability Calculations
Basic Probabilities
- P(F): 400/600 = 2/3
- P(F’): 200/600 = 1/3
- P(x): 100/600 = 1/6
- P(x’): 500/600 = 5/6
Intersections
- P(F ∩ x): 40/600 = 0.067
- P(F’ ∩ x): 60/600 = 0.1
- P(F ∩ x’): 360/600 = 0.6
- P(F’ ∩ x’): 140/600 = 0.233
Unions
- P(F ∪ x): P(F) + P(x) – P(F ∩ x)
= 400/600 + 100/600 – 40/600 = 0.7666 - P(F’ ∪ x): P(F’) + P(x) – P(F’ ∩ x)
= 200/600 + 100/600 – 60/600 = 0.400 - P(F ∪ x’): P(F) + P(x’) – P(F ∩ x’)
= 400/600 + 500/600 – 360/600 = 0.900 - P(F’ ∪ x’): P(F’) + P(x’) – P(F’ ∩ x’)
= 200/600 + 500/600 – 140/600 = 0.9333
Conditional Probabilities
- P(F | x): P(F ∩ x) / P(x)
= 40/100 = 0.4000 - P(F’ | x): P(F’ ∩ x) / P(x)
= 60/100 = 0.6000 - P(F | x’): P(F ∩ x’) / P(x’)
= 360/500 = 0.7200 - P(F’ | x’): P(F’ ∩ x’) / P(x’)
= 140/500 = 0.2800 - P(x | F): P(F ∩ x) / P(F)
= 40/400 = 0.1000 - P(x’ | F): P(F ∩ x’) / P(F)
= 360/400 = 0.9000 - P(x | F’): P(F’ ∩ x) / P(F’)
= 60/200 = 0.3000 - P(x’ | F’): P(F’ ∩ x’) / P(F’)
= 140/200 = 0.7000
Probability of Selecting Defective Fuses
Suppose that we have a fuse box containing 20 fuses, of which 5 are defective. If 2 fuses are selected at random and removed from the box in succession without replacing the first, what is the probability that both fuses are defective?
The probability that the first fuse is defective is 5/20. Since the first fuse is removed, the probability that the second fuse is also defective is 4/19 (as there are now 4 defective fuses left out of 19 total). Thus, the probability of both fuses being defective is:
P(D1∩D2)=520×419=595P(D_1 ∩ D_2) = 5/20*4/19*=5/95
Law of Probability for Dependent Events
The law of probability states that the probability of two dependent events occurring together is the product of the probabilities of the individual events, adjusted for dependency. In this case, the selection of the second fuse depends on the outcome of the first selection. The probability of selecting the first defective fuse is 5/20, and given that the first one was defective, the probability of selecting a second defective fuse is 4/19. Therefore, the probability of both being defective is 5/95, demonstrating the multiplication rule for dependent events.
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