destinyterrea
05.02.2020 •
Mathematics
Aroom has 3 lamps. from a collection of 12 light bulbs of which 5 are not working, richard selects 3 at random and puts them in the sockets. what is the probability that he will have light?
Solved
Show answers
More tips
- S Style and Beauty How to knit a hooded cowl?...
- H Horoscopes, Magic, Divination How to Cast a Love Spell on a Guy? Guide for Guys...
- S Style and Beauty Tricks and Tips: How to Get Rid of Freckles...
- H Health and Medicine How to perform artificial respiration?...
- C Computers and Internet How to Get Rid of Windows Genuine Check?...
- F Food and Cooking The Disease That Haunted Abraham Lincoln...
- S Style and Beauty How to Make Your Lips Fuller? Ideas and Tips for Beautiful Lips...
- S Style and Beauty How are artificial nails removed?...
- F Family and Home How to Sew Curtain Tapes: Best Tips from Professionals...
- F Family and Home How to Properly Use a Water Level?...
Answers on questions: Mathematics
- M Mathematics Use the graph of the line to answer each question....
- M Mathematics Prove (1/(1+cosx)) - (1/(1-cosx) 1 ; 1 - (1 + cosx) (1 - cosx)...
- M Mathematics If there are 16 cups in 1 gallon, how many cups are there in 3 gallons...
- M Mathematics Ihave been stuck on these 2 questions for 10 minutes!...
- M Mathematics What is a solution to —2(5x+2x)—3(5)=—10+2(—7x)—5...
- M Mathematics Pls me asap it needs to be turned in today pls...
- M Mathematics Which are correct representations of the inequality 6x 3+ 4(2x - 1)? select three options.12 2x6x2 3 + 8x - 4-1.5 -1 -0.50 0....
- M Mathematics Given: bisects ∠mrq; ∠rms ≅ ∠rqs triangles r m s and r q s share side r s. point n is on line r s. lines are drawn from point m to point n and from point q to point...
- M Mathematics Jamie and stella are saving money to sign up for a school trip to washington, d.c. in order to sign up for the trip, they must pay $600 upfront. jamie earns his...
- M Mathematics Simplify. (3x2−2x+2)−(x2+5x−5) a.4x^2+3x−3 b.2x^2+3x−3 c.2x^2−7x+7 d.2x^2−3x−3...
Ответ:
(7∗6∗5)÷(12∗11∗10)=.159
or 15.9%
But that's for selecting 3 good bulbs. Do the same with two bulbs
(7∗6)÷(12∗11)=
or 31.8%
And with one bulb
7÷12=.583
or 58.3%So the answer seems to depend on the criteria for "having light", which is what makes it so confusing. Plus, we'd like to consider the probability of any of these conditions obtaining in any combination. The process yields errors if we keep trying to figure out how to choose a good bulb.
So let's turn the question on its head and ask what's the probability of no light?
( 12
3 )=12!÷3!9!=20
then how many ways to choose 3 bad bulbs?
(12
5 )=12!÷5!7!=792
Divide and subtract from one to get the probability of there being light. 1 - 20/792 = 97.5%
Ответ:
idk :(
Step-by-step explanation: