In: Statistics and Probability
There are 10 forms on a desk. 6 of these are Type A forms, and the other four are Type B forms.
Suppose only 4 forms can be processed in one day. If these four are randomly selected one by one, what is the probability that:
a) exactly two forms from each of the two types are selected for
processing?
b) each succeeding form is of a different type from its
precessor
If 6 of these forms can be selected in one day,
(c) what is the probability that only one of the two types of forms remains on his desk?
a) from hypergeometric distribution probability that exactly two forms from each of the two types are selected for processing = =0.4286
b)total numebr of arrangent =N(all 4 are type A +3 of type A and 1 of type B+2 of type A and 2 of type B+1 of type A and 3 of type B+0 of type A and 4 of type B)
=4!/(0!*4!)+4!/(1!*3!)+4!/(2!*2!)+4!/(3!*1!)+4!/(4!*0!) =16
numebr of ways to arrnage forms so that each succeeding form is of a different type from its precessor
=N(ABAB or BABA) =2
hence probability =2/16 =1/8 =0.125
c)
probability that only one of the two types of forms remains on his desk
=P(sleecting 2 of form A and 4 of form B+selecting all 6 of form A)
= =0.0714+0.0048 =0.0762