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

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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 $(c)$.

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

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