Following is a question from my textbook. My approach is different from one explained in the book. I cannot understand what is wrong with my solution. I have explained both solutions below. Kindly clarify.
Qst: A sample consists of $10$ items. Let $X$ denote the number of non detectives. Test $H_0:X=5$ vs $H_1:X=4$. The sample is rejected if $2$ items drawn from this sample randomly with replacement is defective. Find the size of the test.
Text book solution: they used the simple probability and got the solution as $5/10*5/10 =0.25$.
My solution: I considered this as as a situation following the binomial distribution. Sample is rejected if #defectives$=2$. Under $H_0$, $P($item is defective$)=5/10=0.5$ Hence, size= 10C2$* (0.5)^{10}=0.04$.
I am confused. Kindly help me identify where I have gone wrong.
If you imagine the 10 items as 10 faces on a die, then you will see that you are rolling twice and that in both case you must get a defective items. So the correct counting factor is $2C2$, not $10C2$.
Addendum: I misread your answer. You made more mistakes.
I agree that you can consider this as a binomial distribution, however, you only have $n=2$ independent trials (since you draw twice) with probability $p = 5/10$ of success. Thus, the correct probability is $$\binom{2}{2}\left(\frac{5}{10}\right)^2\left(\frac{5}{10}\right)^0 = \left(\frac{5}{10}\right)^2.$$