How many arrangements are possible using the letters in the word FUZZY if each letter “Z” is distinctly different than the other? How many arrangements are possible if the letter “Z” is interchangeable with the other? help i dont know how to do this .can someone please explain step by step how to do this problem to me

Question

jim_thompson5910 · Answer

You have 6 letters, so there are 6! = 720 different ways to arrange them if the Z's are distinguishable

jim_thompson5910 · Answer

If they are not distinguishable, then 720 is too high of a count and you have to divide by 2! = 2 to correct for this over-counting

anonymous · Answer

i just have a ? how did you get that there are 6 letters?

jim_thompson5910 · Answer

FUZZY has 6 letters

jim_thompson5910 · Answer

oh wait...lol miscounted...it should be 5

jim_thompson5910 · Answer

looked at it too quickly

jim_thompson5910 · Answer

so it would be 5! = 120

and 120/2 = 60

anonymous · Answer

thank you thats what i thought  lol but i didnt know if there was more to it ......

jim_thompson5910 · Answer

yw

anonymous · Answer

could you help me with another ?

jim_thompson5910 · Answer

sure

anonymous · Answer

the police department is keeping track of  distracted drivers and accidents. they have found that if a driver is distracted, the driver has a 30% chance of being in an accident. if the driver is not distracted, the driver has a 2% chance of being in an accident. the probability of a driver being distracted is 10%. if needed, create a tree diagram. what is the probability a driver will be in an accident? explain. 2. what is the probability that a driver who was in an accident was distracted? explain.

jim_thompson5910 · Answer

ok one sec

jim_thompson5910 · Answer

ok so we'll have this tree diagram

|dw:1363741727389:dw|

jim_thompson5910 · Answer

if you multiply out each branch, you'll get this

|dw:1363741876350:dw|

jim_thompson5910 · Answer

so the probability of getting into an accident is

0.03 + 0.018 = 0.048

this is if you're distracted or not distracted

jim_thompson5910 · Answer

So we want to find

P(distracted|accident)

-------------------------------------------------------

P(distracted|accident) = P(distracted and accident)/P(accident)

P(distracted|accident) = 0.03/0.048

P(distracted|accident) = 0.625

which is a 62.5% chance

anonymous · Answer

thank you very much now i get it.

jim_thompson5910 · Answer

yw