Brian has a bag that contains 14 red marbles and 12 yellow marbles. He selects a marble at random, and then, without replacing the first one, selects another marble at random. What is the probability that Brian selects a red marble and then a yellow marble? Round your answer to the nearest percent. P(red and yellow) ≈ ________ %
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Follow these steps: A) Number of red marbles = ? B) Total number of marbles before first drawing = ? C) Divide number in A) by number in B) D) Number of yellow marbles = ? E) Total number of marbles left after first drawing = ? F) Divide number in D) by number in E) G) Multiply number in C) by number in F) H) Your solution is the product in G)
i am not sure, i need help, i will give medla and fan
When you do , The correct answer is red = 14 marbles = 56%, then yellow = 12 marbles = 46%
ok
step e is confusing
let's go through it step by step we have 14 red marbles. How many marbles are there total?
26
for step c i got 1.85 (and a whole bunch of other random numbers)
and for step d i got 12
so the probability of drawing red is 14/26 I divided the number of red by the number total
i got 7/13
after you draw a red marble, how many total marbles are left?
if u draw 1, that would be 25
so what is the probability of getting yellow?
is it 25?
you represent it as a fraction (see how I did red above)
25/26
how many yellow?
right?
there is 12 yellow
i simplified 25/26 and got 0.961538
that is in decimal
in percent it is...
so the probability of picking yellow is 12/25
first draw: probability of picking red = 14/26 second draw: probability of picking yellow = 12/25
multiply them together (14/26)*(12/25) = 0.25846153846153
okay... i am following u
then just convert 0.25846153846153 to a percent and you're done
25.846153846%
that is what I got @jim_thompson5910
am i right?
which is rounded to 26
yeah 26% looks good
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