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DEEBA
 2 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
 2 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

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

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

agent0smith
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1use a tree diagram, i'll draw one

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

agent0smith
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1so basically in the end, you did need to multiply it by 2 ;)

agent0smith
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1right, no replacement! Sorry, it's been a while

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

agent0smith
 2 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%.
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