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Mathematics 9 Online
OpenStudy (pulsified333):

Assume that each bag contains 5 balls. Bag a contains 1 red and 4 white, while bag b contains 1 red, 2 white, and 2 blue. You randomly select one ball from bag a, note the color, and place the ball in bag b. You then select a ball from bag b at random and make note of its color. (1) What is the probability that both balls are red? (2) What is the probability that both balls are red given that the first ball you drew was red?

OpenStudy (anonymous):

you got a lot of work to do

OpenStudy (anonymous):

oh nvm the first one is not that hard at all

OpenStudy (anonymous):

what is the probably the first ball is red?

OpenStudy (anonymous):

Chance of getting a red ball from bag A --> 3/5 Place red ball into bag B --> Bag B now has 6 balls, 2 of which are red. Chance of getting a red ball from bag B --> 2/6 = 1/3 Chance of both events being true --> (3/5) * (1/3) = 1/5 I'm not sure if that's right.

OpenStudy (pulsified333):

yeah I know. As you might expect my brain isn't functioning at its best right now but I am trying

OpenStudy (anonymous):

@SataniCross idea is right, you number is wrong

OpenStudy (anonymous):

that's what i thought

OpenStudy (anonymous):

only the first number you put \(\frac{3}{5}\) but it should be \(\frac{1}{5}\)

OpenStudy (anonymous):

you got this? or you want to walk through it?

OpenStudy (pulsified333):

would 1 be 1/15?

OpenStudy (anonymous):

\[\frac{1}{5}\times \frac{2}{6}\] yeah you are n right

OpenStudy (pulsified333):

then for 2, Would it be (1/15)/(1/5)?

OpenStudy (anonymous):

yes

OpenStudy (pulsified333):

:D thanks

OpenStudy (anonymous):

yw you did all the work

OpenStudy (pulsified333):

yeah but you helped with explaining the technique on the previous problem :D

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