## 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

1. Chelsea04

1/8*1/9 so 1/72

2. Chelsea04

you find the probability of getting a hard centred: 1/8 then the probability of getting a soft centred: 1/9 then multiply them together

3. Chelsea04

it's wrong isn't it

4. agent0smith

I think you may have to double your answer @Chelsea04 as you could get one hard, then one soft, or one soft, then one hard.

5. Chelsea04

i'm thinking of a different question. oh, right, yea. that i think makes much more sense

6. Chelsea04

no! no! I got it now! hard centred: 8/17 soft centred: 9/17 multiply these together 72/289

7. agent0smith

Yeah that sounds a bit better... been a while since i've done one of these.

8. Chelsea04

do you multiply that by 2?

9. Chelsea04

yea, me too! been like 3 months

10. agent0smith

Hmm i'm not sure if that's correct. It may be easier to do it using combinations...

11. Chelsea04

use a tree diagram, i'll draw one

12. agent0smith

There's four possible outcomes: (h, h), (s,s), (h,s), (s,h)

13. agent0smith

(note not all are equally likely as there's 8 hard and 9 soft)

14. agent0smith

$\left(\frac{ 8 }{ 17 } \times \frac{ 9 }{ 16 }\right) + \left(\frac{ 9 }{ 17 } \times \frac{ 8 }{ 16 }\right)$ I think this covers all possibilities...

15. Chelsea04

yea, it does

16. Chelsea04

so basically in the end, you did need to multiply it by 2 ;)

17. agent0smith

Yep, but your original equation had them both over 17, not one over 17 and one over 16.

18. Chelsea04

right, no replacement! Sorry, it's been a while

19. DEEBA

is it the rite ans

20. Chelsea04

$\left(\frac{ 8 }{ 17 } \times \frac{ 9 }{ 16 }\right) + \left(\frac{ 9 }{ 17 } \times \frac{ 8 }{ 16 }\right) = 2 \times \left(\frac{ 9 \times 8 }{ 17 \times 16 } \right) = 0.529$ 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%.