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Question: A box \[{B_1}\] contains 1 white ball, 3 red balls and 2 black balls. Another box \[{B_2}\] contains...

A box B1{B_1} contains 1 white ball, 3 red balls and 2 black balls. Another box B2{B_2} contains 2 white balls, 3 red balls and 4 black balls. A third box B3{B_3} contains 3 white balls, 4 red balls and 5 black balls.
If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2{B_2} is,
A.116181\dfrac{{116}}{{181}}
B. 126181\dfrac{{126}}{{181}}
C. 65181\dfrac{{65}}{{181}}
D. 55181\dfrac{{55}}{{181}}

Explanation

Solution

We use the concept of probability to solve for the value. Consider drawing one red and one white ball as even and calculate the probability of selecting one box from 3 total boxes first. Later use the concept of combinations to write probabilities of selecting one red and one white ball from each of the boxes separately. Calculate total number of possible outcomes by multiplying probabilities of selecting that particular box with the respective probability of obtaining a red or a white ball from that box and add the three. Calculate final probability by dividing probability for required box by probability of all boxes.

  • Probability of an event is given by the number of possibilities divided by total number of possibilities.
  • Combination is given by nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}

Complete step-by-step solution:
Here we are given three boxes B1,B2,B3{B_1},{B_2},{B_3}
Probability of selecting 1 box from these 3 boxes is given as P(B1)=13,P(B2)=13,P(B3)=13P({B_1}) = \dfrac{1}{3},P({B_2}) = \dfrac{1}{3},P({B_3}) = \dfrac{1}{3}...............… (1)
Now we write a separate number of balls in each box and calculate the probability of selecting 1 red and 1 white ball from that respective box.

  1. Box B1{B_1}:
    Number of white balls =1 = 1
    Number of red balls =3 = 3
    Number of black balls =2 = 2
    So, total number of balls =6 = 6
    Probability of selecting 1 red ball and 1 white ball from box {B_1}$$$$ = \dfrac{{^1{C_1}{ \times ^3}{C_1}}}{{^6{C_2}}}
    Open using combination formula i.e. nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
    ⇒\RightarrowProbability of selecting 1 red ball and 1 white ball from box {B_1}$$$$ = \dfrac{{\dfrac{{1!}}{{(1 - 1)!1!}} \times \dfrac{{3!}}{{(3 - 1)!1!}}}}{{\dfrac{{6!}}{{(6 - 2)!2!}}}}
    ⇒\RightarrowProbability of selecting 1 red ball and 1 white ball from box {B_1}$$$$ = \dfrac{{\dfrac{{1!}}{{0!1!}} \times \dfrac{{3 \times 2!}}{{2!1!}}}}{{\dfrac{{6 \times 5 \times 4!}}{{4!2!}}}}
    Cancel possible terms from numerator and denominator
    ⇒\RightarrowProbability of selecting 1 red ball and 1 white ball from box {B_1}$$$$ = \dfrac{{1 \times 3}}{{15}}
    ⇒\RightarrowProbability of selecting 1 red ball and 1 white ball from box {B_1}$$$$ = \dfrac{1}{5}
    Then we can calculate probability of selecting 1 red ball and 1 white ball from box B1{B_1}from given 3 boxes say P1{P_1} as 13×15=115\dfrac{1}{3} \times \dfrac{1}{5} = \dfrac{1}{{15}}
    ⇒P1=115\Rightarrow {P_1} = \dfrac{1}{{15}}..............… (2)
  2. Box B2{B_2}:
    Number of white balls =2 = 2
    Number of red balls =3 = 3
    Number of black balls =4 = 4
    So, total number of balls =9 = 9
    Probability of selecting 1 red ball and 1 white ball from box {B_2}$$$$ = \dfrac{{^2{C_1}{ \times ^3}{C_1}}}{{^9{C_2}}}
    Open using combination formula i.e. nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
    ⇒\RightarrowProbability of selecting 1 red ball and 1 white ball from box {B_2}$$$$ = \dfrac{{\dfrac{{2!}}{{(2 - 1)!1!}} \times \dfrac{{3!}}{{(3 - 1)!1!}}}}{{\dfrac{{9!}}{{(9 - 2)!2!}}}}
    ⇒\RightarrowProbability of selecting 1 red ball and 1 white ball from box {B_2}$$$$ = \dfrac{{\dfrac{{2!}}{{1!1!}} \times \dfrac{{3 \times 2!}}{{2!1!}}}}{{\dfrac{{9 \times 8 \times 7!}}{{7!2!}}}}
    Cancel possible terms from numerator and denominator
    ⇒\RightarrowProbability of selecting 1 red ball and 1 white ball from box {B_2}$$$$ = \dfrac{{2 \times 3}}{{36}}
    ⇒\RightarrowProbability of selecting 1 red ball and 1 white ball from box {B_2}$$$$ = \dfrac{1}{6}
    Then we can calculate probability of selecting 1 red ball and 1 white ball from box B2{B_2}from given 3 boxes say P2{P_2} as 13×16=118\dfrac{1}{3} \times \dfrac{1}{6} = \dfrac{1}{{18}}
    ⇒P2=118\Rightarrow {P_2} = \dfrac{1}{{18}}.............… (3)
  3. Box B3{B_3}:
    Number of white balls =3 = 3
    Number of red balls =4 = 4
    Number of black balls =5 = 5
    So, total number of balls =12 = 12
    Probability of selecting 1 red ball and 1 white ball from box {B_3}$$$$ = \dfrac{{^3{C_1}{ \times ^4}{C_1}}}{{^{12}{C_2}}}
    Open using combination formula i.e. nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
    ⇒\RightarrowProbability of selecting 1 red ball and 1 white ball from box {B_3}$$$$ = \dfrac{{\dfrac{{3!}}{{(3 - 1)!1!}} \times \dfrac{{4!}}{{(4 - 1)!1!}}}}{{\dfrac{{12!}}{{(12 - 2)!2!}}}}
    ⇒\RightarrowProbability of selecting 1 red ball and 1 white ball from box {B_3}$$$$ = \dfrac{{\dfrac{{3 \times 2!}}{{2!1!}} \times \dfrac{{4 \times 3!}}{{3!1!}}}}{{\dfrac{{12 \times 11 \times 10!}}{{10!2!}}}}
    Cancel possible terms from numerator and denominator
    ⇒\RightarrowProbability of selecting 1 red ball and 1 white ball from box {B_3}$$$$ = \dfrac{{3 \times 4}}{{66}}
    ⇒\RightarrowProbability of selecting 1 red ball and 1 white ball from box {B_3}$$$$ = \dfrac{2}{{11}}
    Then we can calculate probability of selecting 1 red ball and 1 white ball from box B3{B_3} from given 3 boxes say P3{P_3} as 13×211=233\dfrac{1}{3} \times \dfrac{2}{{11}} = \dfrac{2}{{33}}
    ⇒P3=233\Rightarrow {P_3} = \dfrac{2}{{33}}..............… (4)
    Then the probability of drawing 1 red ball and 1 white ball from box B2{B_2}will be=P2P1+P2+P3 = \dfrac{{{P_2}}}{{{P_1} + {P_2} + {P_3}}}
    Substitute the values from equations (1), (2) and (3)
    ⇒\RightarrowProbability =118115+118+233 = \dfrac{{\dfrac{1}{{18}}}}{{\dfrac{1}{{15}} + \dfrac{1}{{18}} + \dfrac{2}{{33}}}}
    Take LCM in the denominator
    ⇒\RightarrowProbability =11818×33+15×33+2×15×1815×18×33 = \dfrac{{\dfrac{1}{{18}}}}{{\dfrac{{18 \times 33 + 15 \times 33 + 2 \times 15 \times 18}}{{15 \times 18 \times 33}}}}
    ⇒\RightarrowProbability =118×15×18×3318×33+15×33+2×15×18 = \dfrac{1}{{18}} \times \dfrac{{15 \times 18 \times 33}}{{18 \times 33 + 15 \times 33 + 2 \times 15 \times 18}}
    Cancel possible terms from numerator and denominator
    ⇒\RightarrowProbability =15×3318×33+15×33+2×15×18 = \dfrac{{15 \times 33}}{{18 \times 33 + 15 \times 33 + 2 \times 15 \times 18}}
    Take common factors from the denominator
    ⇒\RightarrowProbability =3×3(5×11)3×3(6×11+5×11+2×5×6) = \dfrac{{3 \times 3(5 \times 11)}}{{3 \times 3(6 \times 11 + 5 \times 11 + 2 \times 5 \times 6)}}
    Cancel same terms from numerator and denominator
    ⇒\RightarrowProbability =55(66+55+60) = \dfrac{{55}}{{(66 + 55 + 60)}}
    ⇒\RightarrowProbability =55181 = \dfrac{{55}}{{181}}
    ∴\therefore The probability that the 2 balls are drawn from box B2{B_2}is 55181\dfrac{{55}}{{181}}

∴\therefore Option D is correct.

Note: Many students make mistake of writing the answer as the value obtained from the second part in the solution i.e. the value of selecting 1 red and 1 white ball from box B2{B_2}, keep in mind that is just the probability from that box, we have to find the probability of choosing these balls from the three total boxes. Also, while calculating the probabilities always cancel the possible factorial first so that we don’t have to open all of them.