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Mathematics 8 Online
OpenStudy (faiqraees):

There are three colors: red, green, blue. She places a single row of 8 colors. Each color is equally likely to be picked. Find the probability that there are no tiles of the same color next to each other.

OpenStudy (faiqraees):

@hartnn @Photon336 @jhonyy9 @vishweshshrimali5 @ganeshie8 @Directrix @Isaiah.Feynman @IrishBoy123 @inkyvoyd @Astrophysics @kropot72 @ParthKohli

OpenStudy (faiqraees):

Is this reasoning correct? For first color she can pick anything. So probability for first tile is 1 For 2nd till eighth she can only pick two remaining colors. So probability for all other is 2/3 So the probability is (2/3)^7??

ganeshie8 (ganeshie8):

Looks okay to me

OpenStudy (faiqraees):

Okay

ganeshie8 (ganeshie8):

You could also count the number of wsys

ganeshie8 (ganeshie8):

*ways

OpenStudy (faiqraees):

Is this correct if I just refer to arrangements 3x 2^7

ganeshie8 (ganeshie8):

How many 8 letter words can be formed using 3 different letters ?

ganeshie8 (ganeshie8):

Yes you have it!

OpenStudy (faiqraees):

8P3?

OpenStudy (faiqraees):

Why is 8P3 not correct?

ganeshie8 (ganeshie8):

No It is 3^8

OpenStudy (faiqraees):

OHH 8P3 would be correct if repetition is not allowed right?

ganeshie8 (ganeshie8):

Each space has 3 colors to choose from

OpenStudy (faiqraees):

Yes

OpenStudy (faiqraees):

@ganeshie8 Can you also tell me how to infer the number of arrangements using the probability of occurence of something?

OpenStudy (faiqraees):

Like there was a question which concerned with the word AMAZED. They asked how many arrangements will be there in which e will be between the 2 A's. They answer was somehow to infer that in almost 1 in 3 case e will be between A. And then they multiplied it with the total permutations.

OpenStudy (faiqraees):

The problem is how did they know in 1 in 3 cases e will be between A?

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