Question

S) Suppose that A and B are mutually exclusive events for which P(A) = 0.3 and P(B) = 0.5 What is the probability that (a) either A or B occurs? (b) A occurs and B does not occur? (c) both A and B occur? 4.) A forest contains twenty elk, of which five are captured, tagged and then released. Some time later, four elk are captured from this population. What is the probability that exactly two of these are tagged? What assumptions are you making?

Answer 1

Solution :

3) It is supposed that A and B are mutually exclusive events for which P(A) = 0.3 and P(B) = 0.5.

We have to find the probabilities given below !! The information that we have is :

$P(A\cap B)=0\ \ (A,B\ are\ mutually\ exclusive)\ ;\ P(A)=0.3\ ;\ P(B)=0.5$

(a) P(Either A or B occurs) :

$P(Either\ A\ or\ B\ occurs)=P(A \ or\ B)=P(A\cup B)$

$=P(A)+P(B)-P(A\cap B)\ \ [From\ Theory\ of\ Sets]$

$=P(A)+P(B)\ \ \ [\because ,\ P(A\cap B)=0\ is\ given]$

$=0.3+0.5=0.8$

$\mathbf{P(Either\ A\ or\ B\ occurs)=0.8...................................(Ans)}$

(b) P(A occurs and B does not occur) :

$P(A\ occurs\ and\ B\ does\ not\ occur)=P(A\cap B^c)$

$=P(A)-P(A\cap B)\ \ [From\ Theory\ of\ Sets]$

$=P(A)\ \ \ [\because ,\ P(A\cap B)=0\ is\ given]$

$=0.3$

$\mathbf{P(A\ occurs\ and\ B\ does\ not\ occur)=0.3...................................(Ans)}$

(c) P(Both A and B occur) :

$P(Both\ A\ and\ B\ occur)=P(A\ and\ B)=P(A\cap B)$

$=0\ \ \ [\because ,\ A\ and\ B\ are\ Mutually\ Exclusive]$

$\mathbf{P(A\ and\ B\ occur)=0...................................(Ans)}$

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4) It is given that a forest contain twenty elk, of which five are captured, tagged and then released. Some time later, four elk are captured from this population. We have to find the probability that exactly two of these are tagged and we also have to mention the assumptions that we are making.
We have, Total number of elk in the forest = 20 ; Number of elk captured, tagged and then released = 5 ;

So, according to the problem, we have,

$Total\ number\ of\ events=\binom{20}{4}$

$Total\ number\ of\ ways\ 2\ will\ be\ tagged\ out\ of\ 4=\binom{5}{2}*\binom{15}{2}$

Let "A" be the event denoting exactly 2 elk are tagged out of 4 which are captured from the population.

$P(A)=\frac{Total\ number\ of\ ways\ 2\ will\ be\ tagged\ out\ of\ 4}{Total\ number\ of\ events}$

  $=\frac{\binom{5}{2}*\binom{15}{2}}{\binom{20}{4}}=\frac{10*105}{4845}=0.21672$

Prob. that exactly 2 elk are tagged out of 4 captured from the population = 0.21672...............................(Ans)

The assumptions that are made while calculating the required probability :

1) The number of trials (number of elk captured) is fixed.

2) Each elk in the population has equal probability of being chosen.

3) The trials are independent, i.e., event of choosing one elk does not affect the event of choosing another elk.

4) No elk was born or dead in the period of capturing the elks.

5) The above given event follows a Binomial Distribution.

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