Take the class size as 100 students (probabilities are unaffected by the total). Then 40 have the misconception and 60 do not. Of the 40 with it, 80% write "Bijoy" = 32; of the 60 without it, 10% write "Bijoy" = 6. So 38 students in all answer "Bijoy".

Part (a)

1. P(no misconception but answers Bijoy) = fraction of the whole class in the no-misconception group who chose Bijoy = 6/100 = 0.06.
2. P(a student answers Bijoy) = total Bijoy answers / class = 38/100 = 0.38. …

Setting up the numbers. Assume 100 students. The teacher finds 40% carry the decimal-comparison misconception (reading $47.07$ as "4707" and so calling Bijoy the farthest) and 60% do not. Among the 40 with the misconception, 80% answer "Bijoy": $0.80\times40 = 32$. Among the 60 without it, only 10% answer "Bijoy" (since 90% did not): $0.10\times60 = 6$. Hence the total who answer "Bijoy" is $32+6 = 38$.

Part (a)

(i) A student who does not hold the misconception yet still writes Bijoy: that is the no-misconception group (60% of class) times the 10% of them who chose Bijoy, i.e. $0.60\times0.10 = 0.06$, or 6 students out of 100.

(ii) A randomly picked student answers Bijoy by either route: with-misconception ($0.40\times0.80 = 0.32$) plus without-misconception ($0.60\times0.10 = 0.06$). Total $= 0.32+0.06 = 0.38$.

(iii)(a) Given the student answered Bijoy, the chance they hold the misconception is, by Bayes' theorem,

$P(\text{misconception}\mid\text{Bijoy}) = \frac{0.32}{0.38} = \frac{32}{38} = \frac{16}{19} \approx 0.842.$ …

Method: Reverse (Bayes) probability via a hypothetical population

Use this for "given the effect, find the probability of the cause" questions — you know how likely each group is to produce an observation and must, from that observation, judge which group a person came from.

Steps

Step 1: Split the population into its causes

Identify the mutually exclusive groups (here: "has the misconception" vs "does not") and their prior fractions. Taking a convenient round total such as $100$ people turns every percentage into a whole-number count and removes fractions from the arithmetic.

Step 2: Apply the conditional rates to each branch

For each group, multiply its size by the rate at which it produces the observed answer. This is the multiplication rule

Step 3: Total probability of the observation

Add the branch results to get everyone who gives that answer: …

Common Mistakes

Mistake 1: Answering (iii) with $P(\text{Bijoy}\mid\text{misconception})=0.8$ instead of $P(\text{misconception}\mid\text{Bijoy})$.

Why it's wrong: the question conditions on the answer given, not on the group, so you must reverse the conditioning with Bayes. Correct approach: $P(\text{misconception}\mid\text{Bijoy})=\dfrac{0.32}{0.38}=\dfrac{16}{19}\approx 0.842$.

Mistake 2: Taking $P(\text{Bijoy})=0.32$ and forgetting the no-misconception students who still chose Bijoy. …