Question 1013446: Hello,
I guess I am really lost, I posted a question yesterday and someone was nice enough to post an answer but I don't understand where the numbers came from.
Here was the question:
I have the answers to the following questions but am not sure of the process to solve them. Thanks for the help.
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?
( 391 / 400 )
Here is the answer posted:
The probability that one only hits is (0.55)(0.05)=0.0275+ (0.45)(0.95)=0.4275
They add up to 0.4550, or 45.5%, which is 91/200.
At least one is 1-probability that both miss. That is 0.45*0.05=0.0225
That is 2.25% or 2.25 per 100 or 9 per 400 that both miss. Subtract 9 from 400 and 391/400 is the probability that at least 1 will hit.
Where did .05 and .45 come from?
Thanks.
Answer by macston(5194)   (Show Source):
You can put this solution on YOUR website! .
***- added
.
*** - One hits the target 55 out of 100 times.
*** - The other 45 times he misses. Probability of missing=(45/100=0.45)
*** - The second hits 95 out of 100 times.
*** - The other 5 times he misses. Probability of missing=(5/100=0.05)
.
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?
( 391 / 400 )
Here is the answer posted:
The probability that one only hits is (0.55)(0.05)=0.0275+ (0.45)(0.95)=0.4275
.
*** - This is (probability first hits)*(second misses)+(first misses)*(second hits)
.
They add up to 0.4550, or 45.5%, which is 91/200.
At least one is 1-probability that both miss. That is 0.45*0.05=0.0225
.
*** - This is (probability first misses)*(probability second misses)
.
That is 2.25% or 2.25 per 100 or 9 per 400 that both miss. Subtract 9 from 400 and 391/400 is the probability that at least 1 will hit.
.
*** - You can also calculate this (at least one hits) by:
*** - (probability only one hits)+(probability both hit)
*** - 0.4550+(0.95)(0.55)=0.4550+0.5225=0.9775 probability at least one will hit
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