Question 4 of 8: Probability Distribution of Random-Guess Score

Question 4: Probability Distribution of Random-Guess Score (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Four independent questions guessed at random: two with 4 options (success prob. $\tfrac14$ each) and two with 5 options (success prob. $\tfrac15$ each). $Y$ = total number correct.

Find. The full probability distribution of $Y$ (0–4 correct) and $P(Y\lt 2)$.

Approach. $Y=\sum_{i=1}^4 B_i$ is the sum of four independent but non-identical Bernoulli trials (not a single Binomial, since the success probabilities differ), so build the pmf by convolution — enumerating all $2^4=16$ correct/incorrect outcomes and summing their probabilities by score.

1. Set up the four Bernoulli trials. $B_1,B_2\sim\text{Bernoulli}(0.25)$ (four-option questions), $B_3,B_4\sim\text{Bernoulli}(0.20)$ (five-option questions), all independent, $Y=B_1+B_2+B_3+B_4$.
2. Convolve to get the pmf of Y. Summing the probability of every one of the 16 correct/incorrect combinations by its total score gives $$P(Y=0)=(0.75)^2(0.80)^2=0.3600$$ $$P(Y=1)=0.4200,\quad P(Y=2)=0.1825,\quad P(Y=3)=0.0350,\quad P(Y=4)=0.0025$$ (the five values sum to 1.0000, confirming the pmf is complete).
3. Probability Y is smaller than two. $$P(Y\lt 2)=P(Y=0)+P(Y=1)=0.3600+0.4200$$ $$\boxed{P(Y\lt 2)=0.7800}$$