Since each trial has the same probability of success,
Let, <span><span><span>Xi</span>=1</span></span> if the <span><span>i<span>th</span></span></span> trial is a success (<span>0</span> otherwise). Then, <span><span>X=<span>∑3<span>i=1</span></span><span>Xi</span></span><span>X=<span>∑<span>i=1</span>3</span><span>Xi</span></span></span>,
and <span><span>E[X]=E[<span>∑3<span>i=1</span></span><span>Xi</span>]=<span>∑3<span>i=1</span></span>E[<span>Xi</span>]=<span>∑3<span>i=1</span></span>p=3p=1.8</span><span>E[X]=E[<span>∑<span>i=1</span>3</span><span>Xi</span>]=<span>∑<span>i=1</span>3</span>E[<span>Xi</span>]=<span>∑<span>i=1</span>3</span>p=3p=1.8</span></span>
So, <span><span>p=0.6</span><span>p=0.6</span></span>, and <span><span>P{X=3}=<span>0.63</span></span><span>P{X=3}=<span>0.63</span></span></span>
I thought what I did was sound, but the textbook says the answer to (a) is <span>0.60.6</span> and (b) is <span>00</span>.
Their reasoning (for (a)) is as follows:
