Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

A marble is drawn from a bag containing 3 red marbles and 2 blue marbles and a number cube with faces numbered 1 through 6 is tossed. What is the probability that the marble drawn is blue and the number cube shows a number less than 3? A. 11/15 B. 1/4 C. 1/6 D. 2/15

Miracrown (miracrown):

Lets look at each event first. So first the marble, what is the total number of marbles in the bag?

OpenStudy (anonymous):

Well, that would b 5, right?

Miracrown (miracrown):

Right! So from that how can we try and write or express the probability that the marble will be blue? since there are 2 blue marbles in the bag and 5 in all

OpenStudy (anonymous):

\[\frac{ 2 }{ 5 }\] right?

Miracrown (miracrown):

Yep

Miracrown (miracrown):

\[\frac{ 2 }{ 5 }\] that is for the first event, the marble being blue.

Miracrown (miracrown):

So now we can look at the 2nd event, the number being less than 3 we know there are 6 numbers in all, from 1 to 6, how many of them are less than 3?

OpenStudy (anonymous):

\[\frac{ 2 }{ 6 }\]

Miracrown (miracrown):

Correct! :)

Miracrown (miracrown):

\[\frac{ 2 }{ 5 } \space \space \space \space \space \frac{ 2 }{ 6 }\] That is the probability for our 2nd event. So now how can we try and combine them together? what operation do we need to perform to combine them together, since we need both of them happening together, or at the same time

OpenStudy (anonymous):

We add, right?

Miracrown (miracrown):

That does not work, we need to multiply them --- the operation we need between them is multiplication

OpenStudy (anonymous):

Oops! Sorry

Miracrown (miracrown):

That's alright! No need to be sorry. \[\frac{ 2 }{5 } \space \space \space \times \space \space \space \frac{ 2 }{ 6 } \space = \] Now all you have to do is to just try and simplify that a bit, to get the answer you need

OpenStudy (anonymous):

Is it \[\frac{ 2 }{ 15 }\]

Miracrown (miracrown):

Yep, that's it! :)

OpenStudy (anonymous):

Thx soo much! :D

Miracrown (miracrown):

No worries :D

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!