Applied Mathematics for Electrical Engineering (3130908)

Question-3b-OR

BE | Semester-3 Winter-2019 | 26-11-2019

Q3) (b)

For First Urn

No. of white balls = 10

No. of black balls = 3

For second Urn

No. of white balls = 3

No. of black balls = 5

Now, probability for two balls drawn from the first urn

P (E_{1}) = P (Two white balls) = \frac{{}^{10}C_{2} {}^{3}C_{0}}{{}^{13}C_{2}} = \frac{10 \times 9}{13 \times 12} = \frac{15}{26}

P (E_{2}) = P (Two black balls) = \frac{{}^{10}C_{0} {}^{3}C_{2}}{{}^{13}C_{2}} = \frac{3 \times 2}{13 \times 12} = \frac{1}{26}

P (E_{3}) = P (One white and one black ball) = \frac{{}^{10}C_{1} {}^{3}C_{1}}{{}^{13}C_{2}} = \frac{10 \times 3}{13 \times 12} = \frac{5}{26}

After event $E_{1}$

No. of white balls in second urn = 5

No. of black balls in second urn = 5

After event

E_{2}

No. of white balls in second urn = 3

No. of black balls in second urn = 7

After event

E_{3}

No. of white balls in second urn = 4

No. of black balls in second urn = 6

Let, A=Probability of a ball drawn from the second urn is white.

Then, by conditional probability

P (\frac{A}{E_{1}}) = \frac{5}{10} = \frac{1}{2}

P (\frac{A}{E_{2}}) = \frac{3}{10}

P (\frac{A}{E_{3}}) = \frac{4}{10} = \frac{2}{5}

Therefore, the total probability is

P (A) = P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{A}{E_{2}}) + P (E_{3}) P (\frac{A}{E 3})

P (A) = (\frac{15}{26}) (\frac{1}{2}) + (\frac{1}{26}) (\frac{3}{10}) + (\frac{5}{13}) (\frac{2}{5})

P (A) = \frac{49}{130}