Question 1013462: Hello,
I am studying for a probability test and at this rate I probably will not do well (ha).
I have copied down a few problems with the answers. It would help me a lot if someone could explain how to set up the problem.
Thank you.
question 1:
3/5 of workers in a factory are men, 2/5 are women. 40% of the men and 60% of the women do not smoke. You randomly choose a worker. What is the probability that the chosen worker does not smoke?
( Answer: 12 / 25 )
question 2:
There are 2 jars. In one jar there are 7 black balls and 3 blue balls. In the second jar There are 9 black balls and 6 blue balls. You throw a cube. If you get the number 5 or 6, you choose randomly a ball from the first jar. If you get any other number, you choose randomly from the second jar. What is the probability that you will choose a blue ball from the first jar? What is the probability that you will choose a blue ball?
(Answer 1/10 11/30)
Question 3:
2 people throw a ball at the same time. One of them hits the target 55 out of 100 times, and the other one hits it 95 out of 100 times.
What is the probability that exactly one of the throws Will hit the target?
(answer: 91 / 200 )
What is the probability that at least one of the throws will hit the target?
(answer: 391 / 400 )
Found 2 solutions by Boreal, Theo:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 1. Number of non-smokers are (3/5)(2/5) for men =6/25
and (2/5)(3/5) for women=6/25. They add to 12/25/
2.(1/3) chance of a 5 or 6 *(3/10)=(1/10)
The probability of choosing a blue ball from the second is (2/3)(6/15)=4/15
You add both of those, which is (3/30)+(8/30), and that is (11/30), once I use the same denominators.
3. exactly one throw is is (.55*.05)=0.0275. That is the first alone. The second alone is 0.45*0.95=0.4275. They add to 0.4550, which is 45.5%=91/200.
probability of at least 1 is 1- probability both don't.
They are independent, so probability of both missing is 0.45*0.05=0.0225
1-0.0225=0.9775, and that is 391/400.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! question 1:
3/5 of workers in a factory are men, 2/5 are women. 40% of the men and 60% of the women do not smoke. You randomly choose a worker. What is the probability that the chosen worker does not smoke?
( Answer: 12 / 25 )
answer 1:
3/5 of the workers are men.
40% of them smoke.
.4 * 3/5 = 4/10 * 3/5 = 2/5 * 3/5 = 6/25 of all the workers don't smoke.
they just happen to be all men.
2/5 of the workers are women.
60% of them don't smoke.l
.6 * 2/5 = 6/10 * 2/5 = 3/5 * 2/5 = 6/25 of all the workers don't smoke.
they just happen to be all women.
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to understand how this works, assume there are 100 workers.
3/5 are men so 60 are men.
2/5 are women so 40 are women.
40% of the men don't smoke.
that's .4 * 60 = 24 men who don't smoke.
60% of the women don't smoke.
that's .6 * 40 = 24 women who don't smoke.
24 + 24 = 48 of the workers who don't smoke. 24 of them are men and 24 of them are women.
48/100 = 24/50 = 12/25.
the general formula involved here is p(a and b) = p(a) * p(b)
question 2:
There are 2 jars. In one jar there are 7 black balls and 3 blue balls. In the second jar There are 9 black balls and 6 blue balls. You throw a cube. If you get the number 5 or 6, you choose randomly a ball from the first jar. If you get any other number, you choose randomly from the second jar. What is the probability that you will choose a blue ball from the first jar? What is the probability that you will choose a blue ball?
(Answer 1/10 11/30)
answer 2 part 1:
probability of throwing the dice and getting a 5 or a 6 is 2/6.
probability of getting a blue ball from jar 1 is 3/10.
probability of throwing the dice and getting a 4 or a 5 and then drawing a blue ball from jar 1 is 2/6 * 3/10 = 6/60 = 1/10.
answer 2 part 2:
probability of throwing the dice and getting a 1,2,3, or 4 is 4/6.
probability of getting a blue ball from jar 2 is 6/15.
probability of throwing the dice and getting a 1,2,3, or 4 and then drawing a blue ball from jar 2 is 4/6 * 6/15 = 24/90 = 8/30.
answer 2 part 3:
probability of getting part 1 or getting part 2 is 1/10 + 8/30 = 3/30 + 8/30 = 11/30.
Question 3:
2 people throw a ball at the same time. One of them hits the target 55 out of 100 times, and the other one hits it 95 out of 100 times.
What is the probability that exactly one of the throws will hit the target?
(answer: 91 / 200 )
answer 3 part 1:
p(a or b) = p(a) + p(b) - p(ab)
this give you the probability that a or b or both will score.
to find just a will score or b will score, but not both, you have to subtract another p(ab).
therefore the formula becomes p(a or b but not both) = p(a) + p(b) - 2 * p(ab).
ab means a and b.
so you get p(a) = 55/100 and p(b) = 95/100
p(ab) = p(a and b) = 55/100 * 95/100 = 209/400.
2 * p(ab) = 2 * 209/400 = 209/200.
p(a) + p(b) - 2 * p(ab) = 55/100 + 95/100 - 209/200.
simplify to get 150/100 - 209/200
convert to common denominator of 200 to get 300/200 - 209/200 = 91/200.
question 3 part 2:
What is the probability that at least one of the throws will hit the target?
(answer: 391 / 400 )
answer 3 part 2:
this includes the probability that both will hit the target as well.
therefore, instead of subtracting 2 * p(ab), just subrract p(ab).
you get p(a or b including both) = p(a) + p(b) - p(ab).
this is equal to 55/100 + 95/100 - 209/400
simplify to get 150/100 - 209/400.
convert to common denominator of 400 to get 600/400 - 209/400
combine like terms to get p(a or b including both) = 391/400.
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