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pooja123

  • 3 years ago

a bag contains 3white and 4red balls.two draws of one ball each are made without replacement.what is the probability that one is red and other is white

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  1. RadEn
    • 3 years ago
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    Probability one is red of the first draw is 4/7 Probability one is white of the 2nd draw is 3/6 =1/2 multiply of both up, we get 4/7 * 1/2 = 4/14 = 2/7

  2. RadEn
    • 3 years ago
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    does that make sense, @pooja123 ?

  3. pooja123
    • 3 years ago
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    the answer is 4/7.

  4. RadEn
    • 3 years ago
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    4/7 is just for the probalility of one red only :) do u want draw 2 balls, right ?

  5. pooja123
    • 3 years ago
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    yes they used the formula like 2into4into3/7into 6

  6. RadEn
    • 3 years ago
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    hmm. i dont understand that, actually the formula to get the probability of an event using : P(A) = number of event A / number all event

  7. nubeer
    • 3 years ago
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    hmm well she is right the answer would be 4/7

  8. RadEn
    • 3 years ago
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    why, can u explain more, nubber ?

  9. nubeer
    • 3 years ago
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    we found the probability 2/7 when first red and second white so P[RW] = 2/7 this is one case another possibility is that first is white and second is red P[WR] = 3/7 * 4/6 = 2/7 so total probability of this event = p[WR]+p[RW]

  10. pooja123
    • 3 years ago
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    thanks

  11. RadEn
    • 3 years ago
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    wait, is draw it one by one Or 2 balls direct ?

  12. nubeer
    • 3 years ago
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    hmm i think one by one...

  13. RadEn
    • 3 years ago
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    hmmmm.. yeah, i see now. the question doesnt says the the first draw must red or white, thanks for your corrections @nubeer and sorry for @pooja123 :)

  14. nubeer
    • 3 years ago
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    glad to help :)

  15. pooja123
    • 3 years ago
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    thank u guys

  16. RadEn
    • 3 years ago
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    im not becareful with this question :)

  17. pooja123
    • 3 years ago
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    k

  18. 1411akshay
    • 3 years ago
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    can u see my isosceles triangle solution for one sec

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