Question

many choices? the 9a) Thiny -eight percent of the houses in a cetain igthhood ead the New Yoker, 41 pencent read the What peruent road cxatly one İOb ) What is the probablity ofgetngatleaศ one 12 in twenty-fournhofa.puir ofdice? 1la) Three machine produce widgets. The fiest one podaces 20% of the total output, the second 39% and the thind the otal outpat is defective s are, what percent 11h.) Given that s randomly selested wldget is defoctive, what is the prohablity it came om machine one 12h.)What is Fix, the oumlative istribution of X7 given it has teo chi 13h ) What is the prohability a couple has at let one g a matigle choice tenr with the possible mwwers for each of the 12gstisns whar ihe probabiliby er average, how many can he espect to get righr? w hat is the standard deviation of X, the sumber 14h ) 0s a

Answer 1

8 a ) Total number of ways , 5 persons can get into 15 hotels are (15)⁵ =759375

The first person has 15 choices

The second person has 14 choices

continuing like this ,

Favorable number of ways for 5 persons to get into five different hotels are 15*14*13*12*11 = 360360

Probability that 5 persons to get into five different hotels = 360360 / 759375

= 0.4745

b)

Total number of ways , 5 persons can get into 15 hotels are (15)⁵ =759375

The first person has 15 choices

So the second , third , fourth , fifth person has only one choice( The hotel that first person got into)

Probability that 5 persons to get into same  hotel = 15/ 759375

= 0.00001975

9 a) P(NY) =0.38

P(Ph) = 0.41

P(NY $\bigcap$ Ph) = 0.07

P( NY U Ph) = P(NY) + P(Ph) -P(NY $\bigcap$ Ph)

= 0.38+0.41 -0.07

= 0.72

Therefore , 72 % read at least one of the publication

b) P(only NY) = 0.38-0.07 = 0.31

P(only Ph) = 0.41-0.07 = 0.34

P (exactly one ) = P(only NY or only Ph)

= P( only NY) + P(only Ph)

= 0.65

65% read exactly one publication .