8. A circuit board requires the use of ve dierent 105 that all in the same package form. Unfortunately, when the ve 105 are given to be placed on the board, the IC identication indicating which 10 is which is unreadable. The circuit board assembler thus places the ICs in the ve sockets (A, B, C, D, and E) arbitrarily hoping for the best. (a) What is the probability that (i) at least socket A gets the correct 10; (ii) at least socket A and socket B get the proper 10; (iii) at least one socket gets the correct 10; and (iv) no 10 is placed in the correct socket? {13) Find an expression for the probabilities of the events in (ii) and (iv) above if there are 1: I05 and 11. places on the circuit board. Hint: What is the probability that m. specically identied 105 are inserted in their proper socket? If A.- is the event that [C i is inserted in its socket (and we don't care where the other ICs end up), then the event in case (iii) is the union of all the n di'erent events .244- (but note that they are not disjoint events). We have a formula to compute the probability of the union of a number of arbitrary events. There are 5! ways to fill 5 sockets with 5 distinct ICs. There are 4! ways that socket A can be correctly filled, and 3! that both sockets A and B can be correctly filled. Thus (i) P(socket A correctly filled) = 4!/5! = (ii) P(sockets A an B correctly filled) = 3!/5! = 20. To answer (iii), we may proceed as follows: The number of ways that k identified sockets can be correctly filled is clearly (5 - k)!. Let A be the event where the ith socket gets the correct IC. Clearly P(A;) = (same answer as in (i)) for all i, and P(A; n Aj) = 20 (same answer as in (ii) ) for all i, j, i # j [there are () = 10 such cases]. Likewise P(A; n A, n Ax) = 2!/5! = for all i, j, k when they are all different [there are () = 10 such cases], P(A; n A; n Ax n A() = 1!/5! = 300 for all i, j, k, I when they are all different [there are () = 5 such cases], and P(AinAn A3 nAinAs) = 300 The probability that at least one IC is in it correct socket is 5 5 P(JA.) = EP(A.) - EP(AinA;) + > P(AinA; nAk )+ i=1 i=1 1.j. k icj - P(AinA; nAnAl) + P(AinAnAnAnAs) i.j.k.I = 5 x - - 10 x 3 + 10 X 8 - 5 x do 120 + 120 30 19 ~ 0.6333 the answer to (iii)]. Thus the probability that no socket gets the correct IC is 1 - P(at least one socket has corr. IC) = 1 ~ 0.3667 the answer to (iv)]