A community for students.
Here's the question you clicked on:
 0 viewing
Kaederfds
 2 years ago
A bag contains 15 balls of which x balls are green. If 13 more green balls are added to the bag, the probability of drawing a green ball at random from the bag is increased by 1/15. Find the value of x.
Kaederfds
 2 years ago
A bag contains 15 balls of which x balls are green. If 13 more green balls are added to the bag, the probability of drawing a green ball at random from the bag is increased by 1/15. Find the value of x.

This Question is Closed

AndrewNJ
 2 years ago
Best ResponseYou've already chosen the best response.1initial p for pick a green p=x/15 after probability p' is increased by 1/15 so (x/15)+(1/15)=(13+x)/(15+13) (x+1)/15=(13+x)/28 multiply 15+28 in each side so we can have 28(x+1)=15(13+x) 13x=167 \[x \approx 13\]

AndrewNJ
 2 years ago
Best ResponseYou've already chosen the best response.1oh, multiply 15*28 in each side

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0there were originally x green and 15 not green therefore probability is \[\frac{x}{x+15}\] when 13 is added then green becomes x+ 13 and 15 still not green therefore the probability now is \[\frac{x+13}{x+ 13 + 15}\] it says that the new probability is the old probability plus 1/15 therefore... \[\frac{x}{x+15} + \frac{1}{15} = \frac{x+13}{x+13+15}\]

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0do you get this @Kaederfds ??

Kaederfds
 2 years ago
Best ResponseYou've already chosen the best response.0i got 9 what about you?

AndrewNJ
 2 years ago
Best ResponseYou've already chosen the best response.1no! the x represent the green balls but green balls is in 15 balls!

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0thank you for pointing that out @AndrewNJ

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0then it shall be first probability is \[\frac{x}{15}\] then when 13 is added \[\frac{x+13}{28}\] then it says that new probability is old probability plus 1/15 \[\frac{x}{15} + \frac{1}{15} = \frac{x+13}{28}\] since they are common denoms you can add x/15 and 1/15 \[\frac{x+ 1}{15} = \frac{x+ 13}{28}\] so now you cross multiply \[28(x+1) = 15(x+13)\] distribute... \[28x + 28 = 15x + 195\] thisdoesnt look like it results to a whole number though

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0@ParthKohli your expertise are needed ;)

phi
 2 years ago
Best ResponseYou've already chosen the best response.0Things work out better if instead of If 13 more green balls are added we say 3 more green balls are added. So maybe a typo in the problem, or whoever posed the problem neglected the constraint that the answer have an integer solution.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.