Asked by kennedy
Sharon has 24 colored paper clips in her desk drawer:12 are red, 8 are yellow, and 4 are blue. She also has 28 pushpins in her drawer: 11 are white, 14 are orange, and 3 are green.
1. If she picks one paper clip, what is the probability that it will be either blue or yellow.4/24*8/24=1/2
2. If she picks one paper clip and one push pin,what is the probability of picking a red paper clip and a push pin?12/24*11/28
3. If she picks two pushpins with replacement, what is the probability. of picking a green pushpin first and an oranges push pin second?
1. If she picks one paper clip, what is the probability that it will be either blue or yellow.4/24*8/24=1/2
2. If she picks one paper clip and one push pin,what is the probability of picking a red paper clip and a push pin?12/24*11/28
3. If she picks two pushpins with replacement, what is the probability. of picking a green pushpin first and an oranges push pin second?
Answers
Answered by
Reiny
1. No, she picks only one clip
If we consider only the clips, there are 12 clips that are either blue or yellow, so the
prob(blue or yellow) = 12/24 = 1/2
I have ignored the pushpins, since I assume she is able to distinguish between paper clips and pushpins when randomly reaching in
I we allow that she may also choose a pushpin, then the above proability would be 12/52 =3/13
2. could be (red clip then push) or (push then red pin)
= (12/52)(28/51) + (28/52)(12/51)
= 2(28/221) = 56/221
3. prob( green, then orange pushpin) = (3/52)(14/52)
= 21/1352
The wording could be clearer.
e.g. in #2, your answer implies that you want a red clip and a red pushpin, and we would be able to distinguish between clips and pushpins in her desk
My answers reflect the second condition I stated in #1
If we consider only the clips, there are 12 clips that are either blue or yellow, so the
prob(blue or yellow) = 12/24 = 1/2
I have ignored the pushpins, since I assume she is able to distinguish between paper clips and pushpins when randomly reaching in
I we allow that she may also choose a pushpin, then the above proability would be 12/52 =3/13
2. could be (red clip then push) or (push then red pin)
= (12/52)(28/51) + (28/52)(12/51)
= 2(28/221) = 56/221
3. prob( green, then orange pushpin) = (3/52)(14/52)
= 21/1352
The wording could be clearer.
e.g. in #2, your answer implies that you want a red clip and a red pushpin, and we would be able to distinguish between clips and pushpins in her desk
My answers reflect the second condition I stated in #1
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