I In the layout of an oriented circuit board for an electron

I- In the layout of an oriented circuit board for an electronic product, there are 10 different locations for chips. (a) If 5 different types of chips are to be placed on the board, how many different layouts are possible? (b) If the 5 chips that are placed on the board are the same type, how many different layouts are possible? (c) If you have 10 chips each 5 of them are similar (i.e. 2 groups of chips each contains 5 similar chips) to be placed on the board, how many different layouts are possible? 2- For the circuit shown below, what is the probability of having a working path between input and output? (The shown numbers are probability of working for components.) 3- In a football game, statistics about players scoring goals shows the following results. Scoring Not scoring Younger than 30 years (60%) 65% 35% Older than 30 years (40%) 45% 55% (a) What is the probability that a player scores given that he/she is younger than 30 years? (b) If a randomly selected player scored a goal, what is the probability that the player is older than 30 years? (c) For a randomly selected player, what is the overall probability of scoring? 4- Starting from the multiplication rule which states that P(A intersection B)= P(A|B)P(B), prove that if A and B are independent, then P(A intersection B)= P(A)P(B) [Hint: The proof is very short almost one line]

Solution

Solution:

A) To Answer this part we should first know How to place these 5 chips in 10 different place

In this case we have 10 different places or value , each place is chosen from 5 different value (chips here) We shall represents these chips by different names say c1,c2,c3,c4,c5

In this part we can choose any of 5 chips For 1 slot (place ) ,there are 5 different chips We can choose for second place thus we will have 5×5= 25 Possible ways to choose first two places Similary for the remaing places we will have 5^8 or total we can say 5^10

So there will be 5^10 different layouts are possible

= 9765625

B) chips given are of same type

we can assume that same type chips as total one quantity or one unit

now using same concept used in part a we will have

1^10

= 10

c) In this case we have 10 chips divided into 2 groups each having 5 similar chips

we have to groups and 10 places so the no. Of possible layouts are

2^10 =1024