$\begin{array}{1 1}1/2 \\ 1/3 \\ 1/4 \\ 2/3 \end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- The Law of Total Probability relates marginal probabilities to conditional probabilities. The probability of event A is the sum of all conditional probabilities of A given E_n across the sample space, weighted by the proportion of time each occurs, i.e, $P (A) = \sum P (A|E_n) \times P(E_n)$

The urn has 5 red and 5 black balls, i.e, n(R) = 5, n(B) = 5. Total number of balls in the sample space = n(S) = 10.

To calculate the probability that the second ball is red, consider the two cases:

1) The first ball drawn is red $\rightarrow$ 2 more red balls are added $\rightarrow$ The # of red balls in the urn increases, which increases the probability of finding a red ball on the second draw.

P (red ball first) = $\large\frac{n(R)}{n(S)} = \frac{5}{10} = \frac{1}{2}$ = $P(E_1)$

Now, 2 red balls are added $\rightarrow$ n(R) = 7 and n(S) = 12.

P (drawing a red ball now) = $\large \frac {n(R)}{n(S)} = \frac{7}{12}$ =$ P(A|E_1)$

Consider the second case:

2) The first ball is black $\rightarrow$ 2 more black balls are added $\rightarrow$ which increases the total # of balls in the urn, but decreases the probability of finding a red ball next.

P (blackball first) = $\large\frac{n(B)}{n(S)} = \frac{5}{10} = \frac{1}{2}$ = $P(E_2)$

Now, 2 black balls are added $\rightarrow$ n(B) = 7 and n(S) = 12.

P (drawing a red ball now) = $\large \frac {n(R)}{n(S)} = \frac{5}{12}$ =$ P(A|E_2)$

Given the law of total probability, the probability that the 2nd ball is red can be found by summing the conditional probability weighthed by the propotion of time it occurs, i.e.

$P(A) = P (A|E_1) \times P(E_1) + P (A|E_2) \times P(E_2)$

$\Rightarrow P(A) = \large\frac{7}{12} $$\times$ $\large\frac{1}{2}$ +$\large\frac{5}{12}$$\times$$\large\frac{1}{2} = \frac{1}{2}$$\times$$\large\frac{7+5}{12} = \frac{1}{2}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...