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anonymous

  • 4 years ago

Please see the attachment and answer with explanation.

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  1. anonymous
    • 4 years ago
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    Please see these and answer.

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  2. anonymous
    • 4 years ago
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    der?

  3. dumbcow
    • 4 years ago
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    haha thats like a whole assignment on probability please post one question at a time

  4. anonymous
    • 4 years ago
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    lol..yeahh..but they r tiny ones...

  5. anonymous
    • 4 years ago
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    At least solve questions 1, 2 , 3.

  6. vishal_kothari
    • 4 years ago
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    the link is not opening in my computer..so draw the questions...

  7. vishal_kothari
    • 4 years ago
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    draw it...

  8. anonymous
    • 4 years ago
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    wait.

  9. anonymous
    • 4 years ago
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    Question 1 ) From a group of 3 boys and 2 girls, 2 children are selected at random. Find the Sample Space.

  10. vishal_kothari
    • 4 years ago
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    options...

  11. anonymous
    • 4 years ago
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    Options nahin hain. Model paper for SA-2 question.

  12. anonymous
    • 4 years ago
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    Question 2) numbers are thrown getting 2 numbers whose sum is divisible by 4 or 5, is considered as a success. Find the probability og getting success.

  13. vishal_kothari
    • 4 years ago
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    what is tough in that...

  14. anonymous
    • 4 years ago
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    Question1) answer is 20. I guess.

  15. anonymous
    • 4 years ago
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    Question 3) A coin is tossed 3 times. Find the chance that head and tail show alternately. (I think answer is 4/8 = 1/2) Plz verify and say.

  16. anonymous
    • 4 years ago
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    Ok, I'll take a crack at this for you. So, for #1, the Sample Space is defined as the list of all possible outcomes. The question is, is 1 boy and 1 girl the option? Or are they unique, like Boy 1 and Girl 1 vs. Boy 1 and Girl 2 - does that make sense? So, I'm not 100% on how the question is defined to be honest... #2: You need to determine all the possible combinations of the sum of two die (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12), see the ones divisible by 4 or 5, then divide that number by they number of possibilities (in this example there are 11). #3: A coin flip is a 50/50 chance. Alternatively, this written as 1/2 (1 possible successful outcome out of two possible outcomes). So, when flipping a coin three times, you need to multiply the respective probabilities to find your answer. Hope this helps

  17. anonymous
    • 4 years ago
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    Can u please explain questions 1 and 3 in detail? The detailed solution.

  18. anonymous
    • 4 years ago
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    For #1 I really can't - the problem doesn't give enough parameters for me to be sure. The issue is how they define the Sample Space. If one boy and one girl is a possible outcome, is it just ONE possible outcome? Does it matter WHICH boy or girl are chosen? Also, does the order matter? Is one boy and one girl the same or different from one girl and one boy? #3 What exactly are you looking for here? Which part of the explanation do you need help with?

  19. anonymous
    • 4 years ago
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    In question 1, waht matters is the grouping I think. Means Boy-boy, boy0girl, and so on. In question 3, The sample space is: (HHH), (HHT), (HTH), (HTT), (TTT), (TTH), (THT) and (THH). So I think probability of getting head and tail alternately is 4/8 = 1/2.

  20. anonymous
    • 4 years ago
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    For #1, ok, so in that case it would be 20 then? For #3, aren't there only two correct outcomes? THT & HTH?

  21. anonymous
    • 4 years ago
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    THen ok. For no.3, I am doubt. I think u r correct.

  22. anonymous
    • 4 years ago
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    And for #1 - my concern is this, here are the two sample spaces which I can't decide between: BG, BB, GG, GB OR B1G1, B1G2, B2G1, B2G2, B3G1, B3G2, and so on...

  23. anonymous
    • 4 years ago
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    I tried with first assumption though I am not sure.

  24. anonymous
    • 4 years ago
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    Ok sweet, FFM is here - I really want to see his opinion on this

  25. anonymous
    • 4 years ago
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    yeah sure.

  26. anonymous
    • 4 years ago
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    >> From a group of 3 boys and 2 girls, 2 children are selected at random. Find the Sample Space. I will just answer the number in the sample space: \( \binom 32 +\binom 22+ \binom 31\times \binom21 =\cdots\)

  27. anonymous
    • 4 years ago
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    So the final answer? I don't know Permutation and combination those things.

  28. anonymous
    • 4 years ago
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    How could you possible understand probability without understanding elementary combinatorics?

  29. anonymous
    • 4 years ago
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    Those things are not there in our syllabus. We just find the sample space by counting. That's the scope of syllabus till now.

  30. anonymous
    • 4 years ago
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    Please help.

  31. anonymous
    • 4 years ago
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    Hm fair enough, Good luck :)

  32. anonymous
    • 4 years ago
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    bb,gg,bg,gb

  33. anonymous
    • 4 years ago
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    dat was answer for 1st ques..

  34. anonymous
    • 4 years ago
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    But B1G1, B2G2, that can also come na? I did that way. Salini didi said that.

  35. anonymous
    • 4 years ago
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    yeah ...u r right....hwmany marks ques r r these??

  36. anonymous
    • 4 years ago
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    itz....bttr if u write ...as u have written

  37. dumbcow
    • 4 years ago
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    it doesn't indicate that it matters which boy or which girl is selected 4 possibilities GG GB BG BB

  38. anonymous
    • 4 years ago
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    yeah dumbow is also right...

  39. dumbcow
    • 4 years ago
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    and if it did then it would have 20 possibilities 5P2 = 20

  40. anonymous
    • 4 years ago
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    I also got 20 but............

  41. anonymous
    • 4 years ago
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    c the ques . is 2 children are selected ....and children is classified as male female ...so ans is BB GG BG gb

  42. anonymous
    • 4 years ago
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    children are generally clasified as boy girl onle..

  43. anonymous
    • 4 years ago
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    rite rite

  44. anonymous
    • 4 years ago
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    3 ans ..... 2/8 =1/4

  45. dumbcow
    • 4 years ago
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    for #2 does it mean 2 dice are thrown ?

  46. anonymous
    • 4 years ago
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    oops srry mate ...i ead it as dice....

  47. anonymous
    • 4 years ago
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    thanks...

  48. dumbcow
    • 4 years ago
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    i get 4/9 for #2

  49. anonymous
    • 4 years ago
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    24/55

  50. anonymous
    • 4 years ago
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    ?

  51. anonymous
    • 4 years ago
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    did u take 0's into account?/

  52. anonymous
    • 4 years ago
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    der??

  53. dumbcow
    • 4 years ago
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    only numbers divisible be 4or5--> 4,5,8,10,12 3 ways of rolling a 4 4 ways of rolling a 5 5 ways of rolling a 8 3 ways of rolling a 10 1 way of rolling a 12 --> 16/36 = 4/9

  54. anonymous
    • 4 years ago
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    i said na ..its not a dice,.

  55. dumbcow
    • 4 years ago
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    na??

  56. anonymous
    • 4 years ago
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    nope...and second ans is 16/36=4/9

  57. anonymous
    • 4 years ago
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    Thanks a lot

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