Answer to Question #293435 in Statistics and Probability for abeni

Question

WORKOUT PROBLEMS

Instruction: show all the necessary steps for the following problems on the space provided.

2. If two members are chosen at random from a committee consisting of 10 members of which 4 are women. Then what is the probability that the randomly chosen members consist of

A. both women

B. both men.

C. one man and one women

3 The probability that a person stopping at a petrol pump will get his tires checked is 0.12, the probability that he will get his oil checked is 0.29. and the probability that he will get both checked is 0.07.

Answer 1

2.

A.

$$P(WW)=\dfrac{\dbinom{4}{2}\dbinom{10-4}{2-2}}{\dbinom{10}{2}}=\dfrac{6(1)}{45}=\dfrac{2}{15}$$

B.

$$P(MM)=\dfrac{\dbinom{4}{0}\dbinom{10-4}{2}}{\dbinom{10}{2}}=\dfrac{1(15)}{45}=\dfrac{1}{3}$$

C.

$$P(1M\ and \ 1 W)==\dfrac{\dbinom{4}{1}\dbinom{10-4}{2-1}}{\dbinom{10}{2}}=\dfrac{4(6)}{45}=\dfrac{8}{15}$$

3.

Let $A$ denote the event $''$ get tires checked$''.$

Let $B$ denote the event $''$ get oil checked$''.$

The events A and B are independent.

Then

$$P(A\cap B)=P(A)P(B)$$

Given $P(A)=0.12, P(B)=0.29$

$$P(A\cap B)=P(A)P(B)=0.12(0.29)$$

$$=0.0348\not=0.07.$$

The statement that the probability that he will get both checked is 0.07 is False.