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$A: (0.20)(0.20)(0.20)= 0.008$
All Pick B: (0.18)(0.18)(0.18)= 0.005832$B: (0.18)(0.18)(0.18)= 0.005832$
All Pick C: (0.26)(0.26)(0.26)= 0.017576$C: (0.26)(0.26)(0.26)= 0.017576$
All Pick D: (0.32)(0.32)(0.32)= 0.032768$ D: (0.32)(0.32)(0.32)= 0.032768$
All Pick E: (0.04)(0.04)(0.04)=0.000064$E: (0.04)(0.04)(0.04)=0.000064$
Sum= 0.008+0.005832+0.017576+0.032768+0.000064=0.06424Sum$= 0.008+0.005832+0.017576+0.032768+0.000064=0.06424$
Probability= (0.06424) x 100% = 6.42%Probability$= (0.06424) \times 100\% = 6.42\%$
b) (0.04)(0.96)(0.96)= 0.036864$(0.04)(0.96)(0.96)= 0.036864 $
Probability = (0.036864) x 100% = 3.69%$= (0.036864) \times 100\% = 3.69\%$
c) (0.26)(0.26)(0.74) = 0.050024$(0.26)(0.26)(0.74) = 0.050024$
Probability = (0.050024) x 100%= 5.00%$= (0.050024) \times 100\%= 5.00\%$
d) Sum of probability = 1$= 1$
New sum of probability= 1-0.20 = 0.80probability$= 1-0.20 = 0.80$
New probability of B = (0.18/0.80)= 0.225$B = \frac{0.18}{0.80}= 0.225$
All 3$3$ take B$B$: (0.225)3 x 100% = 1.14%$(0.225)^3 \times 100\% = 1.14\%$
