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There are 8 TV sets, 6 are color and 2 are black-white. If 2 [#permalink]
21 Mar 2006, 16:04

There are 8 TV sets, 6 are color and 2 are black-white. If 2 sets are selected at random, what is the probability that at least one of them is color set?

There are 8 TV sets, 6 are color and 2 are black-white. If 2 sets are selected at random, what is the probability that at least one of them is color set?

There are 8 TV sets, 6 are color and 2 are black-white. If 2 sets are selected at random, what is the probability that at least one of them is color set?

What is the prob. that no colour sets are chosen in 2 attempts?
2/8 *1/7= a (too lazy to do the math)

therefore, the prob. to chose at least 2 color sets is 1-a

There are 8 TV sets, 6 are color and 2 are black-white. If 2 sets are selected at random, what is the probability that at least one of them is color set?

1 - (2c2 / 8c2)

= 27/28

i'm going a lil crazy, but what is 2c2 again?? ..

i had a totally different solution _________________

The greater the sacrifice, the greater the Victory

There are 8 TV sets, 6 are color and 2 are black-white. If 2 sets are selected at random, what is the probability that at least one of them is color set?

1 - (2c2 / 8c2)

= 27/28

i'm going a lil crazy, but what is 2c2 again?? ..

i had a totally different solution

2c2 is number of ways to select 2 black/white tv sets out of 2....

There are 8 TV sets, 6 are color and 2 are black-white. If 2 sets are selected at random, what is the probability that at least one of them is color set?

1 - (2c2 / 8c2)

= 27/28

i'm going a lil crazy, but what is 2c2 again?? ..

i had a totally different solution

2c2 is number of ways to select 2 black/white tv sets out of 2....

ok Sima, please bear with me.. can you break the solution down step by step? _________________

The greater the sacrifice, the greater the Victory

First of all, to get the probability that at least one of the TV sets will be color we can use the following formula....

1-(the probability of choosing balck/white tv set both times)

To get that probability we have to calculate total favourable outcomes of choosing 2 black/white tv sets... which is 2c2 since we are calculating the number of ways to choose 2 tv sets out of 2....

... and divide it by the total number of possible outcomes... which is 8c2... since we have to choose 2 tv sets out of 8...

There may be other approaches, what approach did you use?

First of all, to get the probability that at least one of the TV sets will be color we can use the following formula....

1-(the probability of choosing balck/white tv set both times)

To get that probability we have to calculate total favourable outcomes of choosing 2 black/white tv sets... which is 2c2 since we are calculating the number of ways to choose 2 tv sets out of 2....

... and divide it by the total number of possible outcomes... which is 8c2... since we have to choose 2 tv sets out of 8...

There may be other approaches, what approach did you use?

my approach was this:

6 color tv..prob that 1 will be color 6/8, 2 black and white 2/7
6/8*2/7 _________________

The greater the sacrifice, the greater the Victory

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...