From a lot of 10 items containing 3 defectives, a sample of 4 items is drawn at random. Let random variable X denote the number of defective items in the sample.
A. $P(X=1)=\frac{1}{6}$
B. $P(X \leq 1)=\frac{2}{3}$
C. $P(0<X<2)=\frac{1}{2}$
D. $P(X>1)=\frac{1}{30}$
Choose the correct answer from options given below.
Answer & explanation
Correct answer: option 4
$\text{Total items}=10,\;\text{defectives}=3,\;\text{good}=7.$
$\text{Sample size}=4.$
$P(X=x)=\frac{\frac{3!}{x!(3-x)!}\cdot\frac{7!}{(4-x)!(3+x)!}}{\frac{10!}{4!6!}}.$
$P(X=0)=\frac{\frac{7!}{4!3!}}{\frac{10!}{4!6!}}=\frac{35}{210}=\frac{1}{6}.$
$P(X=1)=\frac{3\cdot\frac{7!}{3!4!}}{\frac{10!}{4!6!}}=\frac{105}{210}=\frac{1}{2}.$
$P(X=2)=\frac{3\cdot\frac{7!}{2!5!}}{\frac{10!}{4!6!}}=\frac{63}{210}=\frac{3}{10}.$
$P(X=3)=\frac{\frac{7!}{1!6!}}{\frac{10!}{4!6!}}=\frac{7}{210}=\frac{1}{30}.$
$(A)\;P(X=1)=\frac{1}{2}\neq\frac{1}{6}\;\text{False}.$
$(B)\;P(X\le1)=\frac{1}{6}+\frac{1}{2}=\frac{2}{3}\;\text{True}.$
$(C)\;P(0<\text{ X }<2)=P(X=1)=\frac{1}{2}\;\text{True}.$
$(D)\;P(X>1)=\frac{3}{10}+\frac{1}{30}=\frac{1}{3}\neq\frac{1}{30}\;\text{False}.$
$\text{Correct statements: (B) and (C).}$