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A certain box contains 4 red blocks, 5 blue blocks, and 3 ye
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17 Nov 2019, 04:30
17 Nov 2019, 04:30
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A certain box contains 4 red blocks, 5 blue blocks, and 3 yellow blocks. Judy will select one of these blocks at random from the box, put it back in the box, and then again select a block at random from the box. What is the probability that both of the blocks selected will be yellow?
Give your answer as a fraction.
Kudos for the right answer and solution.
Give your answer as a fraction.
Show: :: OA
\(\frac{1}{16}\)
Kudos for the right answer and solution.
Re: A certain box contains 4 red blocks, 5 blue blocks, and 3 ye
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27 Nov 2019, 07:33
27 Nov 2019, 07:33
The answer is 1/16.
The probability that the selected block is yellow=3/12=1/4. As the 1st block is put back in place, so the probability of drawing a 2nd yellow block=3/12-1/4. So the answer=1/4 X 1/4 = 1/16.
The probability that the selected block is yellow=3/12=1/4. As the 1st block is put back in place, so the probability of drawing a 2nd yellow block=3/12-1/4. So the answer=1/4 X 1/4 = 1/16.
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Re: A certain box contains 4 red blocks, 5 blue blocks, and 3 ye
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08 Dec 2019, 09:45
08 Dec 2019, 09:45
2
Carcass wrote:
A certain box contains 4 red blocks, 5 blue blocks, and 3 yellow blocks. Judy will select one of these blocks at random from the box, put it back in the box, and then again select a block at random from the box. What is the probability that both of the blocks selected will be yellow?
Give your answer as a fraction.
Kudos for the right answer and solution.
Give your answer as a fraction.
Show: :: OA
\(\frac{1}{16}\)
Kudos for the right answer and solution.
4 + 5 + 3 = 12
So there are 12 blocks all together, and 3 of them are yellow
So, P(select yellow block) = 3/12 = 1/4
P(both blocks are yellow) = P(1st block is yellow AND 2nd block is yellow)
= P(1st block is yellow) x P(2nd block is yellow)
= 1/4 x 1/4
= 1/16
ASIDE: Since the first selected block is placed back into the box, the probability of selecting a yellow block first or second will be the same.
Answer: 1/16
Cheers,
Brent
