A community for students.
Here's the question you clicked on:
 0 viewing
DEEBA
 3 years ago
A box has 8 hard centred and 9 soft centred chocolates. two are selected at random.what is the probability that one is hard centred and other is soft centred
DEEBA
 3 years ago
A box has 8 hard centred and 9 soft centred chocolates. two are selected at random.what is the probability that one is hard centred and other is soft centred

This Question is Closed

Chelsea04
 3 years ago
Best ResponseYou've already chosen the best response.1you find the probability of getting a hard centred: 1/8 then the probability of getting a soft centred: 1/9 then multiply them together

agent0smith
 3 years ago
Best ResponseYou've already chosen the best response.1I think you may have to double your answer @Chelsea04 as you could get one hard, then one soft, or one soft, then one hard.

Chelsea04
 3 years ago
Best ResponseYou've already chosen the best response.1i'm thinking of a different question. oh, right, yea. that i think makes much more sense

Chelsea04
 3 years ago
Best ResponseYou've already chosen the best response.1no! no! I got it now! hard centred: 8/17 soft centred: 9/17 multiply these together 72/289

agent0smith
 3 years ago
Best ResponseYou've already chosen the best response.1Yeah that sounds a bit better... been a while since i've done one of these.

Chelsea04
 3 years ago
Best ResponseYou've already chosen the best response.1do you multiply that by 2?

Chelsea04
 3 years ago
Best ResponseYou've already chosen the best response.1yea, me too! been like 3 months

agent0smith
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm i'm not sure if that's correct. It may be easier to do it using combinations...

Chelsea04
 3 years ago
Best ResponseYou've already chosen the best response.1use a tree diagram, i'll draw one

agent0smith
 3 years ago
Best ResponseYou've already chosen the best response.1There's four possible outcomes: (h, h), (s,s), (h,s), (s,h)

agent0smith
 3 years ago
Best ResponseYou've already chosen the best response.1(note not all are equally likely as there's 8 hard and 9 soft)

agent0smith
 3 years ago
Best ResponseYou've already chosen the best response.1\[\left(\frac{ 8 }{ 17 } \times \frac{ 9 }{ 16 }\right) + \left(\frac{ 9 }{ 17 } \times \frac{ 8 }{ 16 }\right) \] I think this covers all possibilities...

Chelsea04
 3 years ago
Best ResponseYou've already chosen the best response.1so basically in the end, you did need to multiply it by 2 ;)

agent0smith
 3 years ago
Best ResponseYou've already chosen the best response.1Yep, but your original equation had them both over 17, not one over 17 and one over 16.

Chelsea04
 3 years ago
Best ResponseYou've already chosen the best response.1right, no replacement! Sorry, it's been a while

Chelsea04
 3 years ago
Best ResponseYou've already chosen the best response.1agent's is the right answer, follow his equation

agent0smith
 3 years ago
Best ResponseYou've already chosen the best response.1\[\left(\frac{ 8 }{ 17 } \times \frac{ 9 }{ 16 }\right) + \left(\frac{ 9 }{ 17 } \times \frac{ 8 }{ 16 }\right) = 2 \times \left(\frac{ 9 \times 8 }{ 17 \times 16 } \right) = 0.529\] So the probability is about 50%, which makes sense given our possible outcomes when picking two chocolates (h for hard, s for soft): (h, h), (s,s), (h,s), (s,h) If all options were equally likely, the probability would be exactly 50%.
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.