A bag contains 3 blue marbles, 9 green marbles and 11 yellow marbles.Twice you draw a marble and replace it. Find p (blue,then green) A.27/529 B.27/23 C.15/259 D.12/23
What is the probability of drawing a blue marble?
So first..there are 3 + 9 + 11 = 23 marbles total right? The probability of drawing 1 blue...is \[\large P = \frac{P_{blue}}{P_{total}} = \frac{3}{23}\] Then it is replaced...so NO marbles are taken out and the second time you go to choose, there are STILL 23 marbles...so when you choose again \[\large P = \frac{P_{green}}{P_{total}} = \frac{9}{23}\] So what is the probability of choosing in this order?
I dont know :?
Well, up to here did all that make sense? probability of choosing blue is the number of blue/ number total And then its replaced so still 23 total marbles And finally probability of green is # of green/ # total All make sense?
yes so far..
Okay good, So, would you agree these are independent events? Does one have ANY effect on the other?
yes
I'm going to assume the "yes" is to the fact that these ARE independent events lol... When you choose the blue and then replace...you have had NO effect on choosing the Green So...to find the probability of 2 independent events...you simply multiply the probabilities So here \[\large P = P(Blue) \times P(Green) = \frac{3}{23} \times \frac{9}{23} = ?\]
so b
Not quite...remember when multiplying *unlike adding* you actually DO the multiplication of the denominator as well :)
yes..
So this would be A because \[\large \frac{3}{23} \times \frac{9}{23} = \frac{3\times 9}{23 \times 23} = \frac{27}{529}\]
ohhhhhhh
Right, also...we could have ruled out B automatically We can never have more than a 100% chance right? Well \(\large \frac{27}{23} = 1.174 \text{ or } 117.4\text{%}\) Impossible :)
Having a probability larger than 1 is equivalent to saying that this event is more certain to happen than a event that must happen, which clearly doesn't make any sense.
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