It is an application of Conditional Probability and Dependent Event.

We have already discussed conditional probability and Independent events.

**Dependent Events :**

Events that are depending on each other, here the sequence in which the events are happening is important. Example:

**let us say a bag has 4 green balls and 7 red balls and we are pulling two balls on after other. **

**What is the probability that both balls are Red?**

**Case 1: Replacement is allowed :**

Solution:

for the first ball: P(red_{1}) = 7 /11

for the second ball :

Since we are putting the first ball then again the total number of ball = 11

So P(red_{2}) = 7/11

So P(red1 and red_{2}) = (7/11) *(7/11) = 49/121

**Case 2: No replacement: **

Here we are not putting the first ball back in the bag.

P(red_{1} ∩^{ }red _{2} )= P(red_{1}) * P(red_{2 }/ red_{1})

= (7/11) * (6/10)

= 42/110

This is the case of dependent events because when you are trying to pull the second ball it depends on the first event.

**Total Probability example 1:-**

let's say we have two bags and bag_{1} contains 4 red and 3 green and bag_{2} contains 5 red and 6 green balls

find out what is the probability of drawing a red ball?

**Solution:- **Now** **drawing a red ball depends on two factors:

1. First, we should know from which bag that ball is drawn and what is the probability of choosing that bag?

2. Composition of that chosen bag?

Let's say the probability of choosing each bag is 1/2

So the probability of drawing a red ball P(red) given as:

**P(red )= Bag **_{1} is selcted and the red ball is drawn OR Bag_{2 }is selcted and red ball is drawn

{ P( Bag _{1} ). (P_{red} /Bag _{1} ) }+ {P(Bag_{2 }). P(P_{red }/ Bag_{2})}

= (1/2)(4/7) + (1/2)(5/11)

=** ( 4/14 ) + (5/22)**