Question

(10) Suppose a drawer contains 5 black socks and 3 white socks. Socks are selected one at a time without replacement until 2 black socks are found. Let X be the number of socks selected. The probability function f(1) = P(X = 2) for X is given by: 1 2 3 4 5 f(x) 0.357 0.357 0.214 0.072

Answer 1

Here, we have to answer 4 parts in this question.The selection of socks is without replacement, hence we use here the concept of Combinations.

The detailed procedure to answer this question is as given in the images below:-

I We are given that a drawer contras 5B (Black) and 3 W (White) socks ie. Total - 5+3 = 8 Socks And socks are selected one at a time. without replacement until 2 Black socks are found. Also -X: No of socks selected And the probability function is- f(x) = P(x-x) for x is given by : If(x) 0.357 0.357 0.214 0.072. Using its the above information we can write cumulative probabilities EXO as follane f(x) 0.357 0.357 0.214 0.072 6.) 0.357 10.714 0.928 1.00 Now, let us answer the questions asked. (a Justification 1 2 -5 as for the range of x given above i beng

Notice that we have 5 B and 3 W = 8 socks So, in order to draw 2B socks, we have to draw at least 2 Socks. Hence the range starts from 2. Further, the number of possible failures = 3 W socks Hence, considering the maximum failures and our successes i. e. 3 W+ 2B - 5 socks. Hence, the range goes upto 5. Hence, the range of x is from 2-5. (b) To Prove :- Why f (2) = P(X= 2) = 0.357 Procedure & Notice that f(2) i.e. P(x=2). I omolier P (No of socks x selected f x=2) So, P (X=2) = n(2 Blacks socks selection choices)! (2 socks selectton choices in Hall f(2)=P (x =2 - 322 - 10 - 0.357 86 28 too +(2) = 0.351

PAGE: DATE: O Graph of Function Cumulative Distribution F(W = P ( xsx). Felag. (C.D.F. 1 2 3 4 5 x=x (No. of socks selected). fig. Graph of Cumulative Distribution Function Here we want to Given that it takes probability that it Black socks obtas the following probabilite more than 2 socks the taker 4 socks to find 2 i e. we have to find P ( x = 41x72) P(X=4X72) = P(X=4 DX22) = P(X=4) P(X72) P(X72) Note that P (X=4) = 0.214 and P(X72) - PCX =2) +P(X24) + P(X=5) - 1-PCX42) P(X72) = -0.357 -0.643. L P (x=4 |×72) = P(x=4) = 0.214 - 0.3328 P(X72) 0.643 Hence, PCX=4]X22) S 0.3328