# How to compute the probability? This is a question from my textbook, there is no answers at the back and I am fairly new to statistics, my answers for questions a-d are below, can anyone check if I answered them correctly?

a) All Pick $$A: (0.20)(0.20)(0.20)= 0.008$$

All Pick $$B: (0.18)(0.18)(0.18)= 0.005832$$

All Pick $$C: (0.26)(0.26)(0.26)= 0.017576$$

All Pick $$D: (0.32)(0.32)(0.32)= 0.032768$$

All Pick $$E: (0.04)(0.04)(0.04)=0.000064$$

Sum$$= 0.008+0.005832+0.017576+0.032768+0.000064=0.06424$$

Probability$$= (0.06424) \times 100\% = 6.42\%$$

b) $$(0.04)(0.96)(0.96)= 0.036864$$

Probability $$= (0.036864) \times 100\% = 3.69\%$$

c) $$(0.26)(0.26)(0.74) = 0.050024$$

Probability $$= (0.050024) \times 100\%= 5.00\%$$

d) Sum of probability $$= 1$$

New sum of probability$$= 1-0.20 = 0.80$$

New probability of $$B = \frac{0.18}{0.80}= 0.225$$

All $$3$$ take $$B$$: $$(0.225)^3 \times 100\% = 1.14\%$$

For part $$(b)$$, you can view it as a Binomial distribution where there are $$3$$ trials and you are looking for the probability that exactly one of them is a success. That is let $$X \sim Bin(3,0.04)$$.

Hence the corresponding probability of

$$P(X=1)= \binom{3}{1}(0.04)(0.96^2)$$

That is your answer has forgotten that we have $$3$$ options to choose the person to choose route $$E$$.

Similarly for part $$(c)$$.

• That makes sense. So I multiplied the probability by 3 for questions b and c to get 11.06% and 15.01% respectively – Timmy Sep 22 '19 at 14:34
• but a and d are fine ? – Timmy Sep 22 '19 at 14:35
• I didn't check the number, but the workings are fine. – Siong Thye Goh Sep 22 '19 at 14:39