Hello everyone, welcome back to another video.
In today's video, we will discuss the questions related to
the topic of probability.
So, let's see what are the questions in those questions.
I hope you have understood all the formulas and theories
that I explained in the last video.
Now, proceeding on to questions.
So, let's see the first question of probability.
So, the question says, balls numbered 1 to 30 are mixed up
and then a ball is drawn at random.
So, what is the experiment in this? You have balls and in
those balls, numbers are written from 1 to 30.
So, this is the experiment.
What is the probability that the ball drawn has a number
which is a multiple of 5 or 7?
So, what we have to do is, we have balls numbered 1 to 30.
We have to pick one ball from those 30 balls.
How many balls do we have to pick?
Eight balls.
So, now we have to see what is the number in that ball
which we have picked.
Okay?
What is the number in that ball which we have picked?
Okay? There must be some number written.
Now, we have to see whether that number is a multiple of 5
or a multiple of 7.
If it is a multiple of any one of these two, then what do
we have to find for that?
We have to find the probability.
So, what is the event?
The event is that we are drawing a ball which has a number
which is a multiple of 5 or 7.
So, first we will see the number of favourable outcomes as
per the event.
So, what is the number of favourable outcomes? Number of
favourable outcomes.
From favourable outcomes,
From where do we see favourable outcomes?
From the sample space, we will see the total number of
elements in the sample space.
From that, we will choose the favourable outcome.
So, what will be the sample space of this?
Sample space S.
In that, we have a ball which has a number 1.
In the ball, we have a number 2.
Similarly, we have all the numbers up to 30 on the ball.
So, how many elements do we have in the sample space?
From 1 to 30, that is 30 elements.
So, what do we have in the sample space S? 30 elements.
Now, from this, we have to choose the favourable outcome.
What should be the number in favourable outcome? That
number should be a multiple of 5 or 7.
So, from 1 to 30,
If we write multiples of 5 or 7, then those will be our
favourable outcomes.
So, what will be the first number? We have 5.
What will be the first multiple of 5? 5 Then, the next
multiple of 5 will be 10.
Then, the next multiple will be 15.
Then, 20.
Then, 25.
And, 30.
Then, we will do the rest.
What will be 35? 35 is bigger than 5.
That means, it is bigger than 30.
There are no balls of that number.
So, we will stop here.
These are all the favourable outcomes of multiples of 5.
Similarly, we will see the multiples of 7.
So, 7 becomes 7.
Then, 14.
Then, 21.
Then, 28.
Then, what will come? 35.
So, 35 becomes bigger than 30. So, we will stop here.
So, how many favourable outcomes do we have? 1, 2, 3, 4, 5,
6, 7, 8, 9, 10.
So, if we have an event,
So, how many favourable outcomes do we have in total? 10.
2, 2, 4, 6, 6, 4, 10. So, we have 10 favourable outcomes in
total.
So, what do we have to do? We have to find the probability.
So, probability of event.
What is event?
That the ball drawn has a number which is a multiple of 5
or 7.
This is the event.
So, write this by referring to E.
So, probability nE by nS.
nE is 10.
nS is 30.
So, this is 1 by 3.
So, the number of probability which is written on the ball
is a multiple of 5 or 7.
That is 1 by 3 probability.
Okay? This was a simple question.
We did not use permutation or combination in this question.
We did not use any other plus or multiply rule.
So, the main thing which I will tell you in the next
questions is how to include permutation and combination in
probability.
We will see that too.
Now, see this.
From a pack of 52 cards.
Whose question is this?
We have a pack of 52 cards.
Two cards are drawn together at random.
So, we have this random event.
Random experiment is of cards.
What is the event?
That the two cards which we are drawing, what is the
probability of both the cards being kings?
So, the two cards which we have drawn.
The first experiment is to draw two cards.
It can be any two cards.
So, the sample space which we will have.
In the sample space, what all elements will we have?
Two cards.
Now, those two cards can be anything.
It can be a king.
It can be of any other color.
So, there can be many possibilities.
Now, should we note down all those possibilities?
Writing and calculating all those possibilities will be
difficult.
If we do it one by one.
So, in this case, what will be our use?
Our use will be of P and C.
That is, of permutation and combination.
So, because we will not have the matter of order in this.
Which card are we choosing first?
And which is the second card?
So, we will apply the formula of combination in this.
So, in the sample space, what will be the members?
What we have to do is,
We have to draw two cards.
So, what do we have in the sample space?
There are 52 cards in total.
We have to draw two cards from 52 cards.
So, from the sample space, we have 52 cards in total.
And from 52, we have to select any two cards.
So, 52C2 will be our NS.
This will become NS.
This will become a number.
That is why it is going to NS.
Because it is giving an element.
We can see which elements are there.
But we should know the number.
That is why we will put it in the formula.
So, what will happen to that? 52 total cards.
And how many we have to draw?
That is two cards.
So, we have to select two from 52.
That is 52C2.
This would be our NS.
This is the total number of possible outcomes.
Now, favorable.
Both cards are kings.
How do we have these two cards?
They are kings.
Now, what are the possibilities of a favorable outcome for
a king?
If both the cards are kings, what will be the probability
of getting the first card?
First, tell me what is the total number of kings?
Total kings.
We have four cards with a total number of kings.
From that, how many we have to draw?
We need two.
We have to select two cards.
So, what is the favorable outcome?
What are you doing?
You are selecting two out of four kings.
So, this will be your 4C2.
Out of four kings, how many we have to draw?
Out of four kings, we have four kings.
We need two.
So, 4C2.
Why are we taking only four?
Why not 52 here?
We have 52 total cards.
We are not taking 52 because we have cards of all choices.
Queen, 2 to 10.
We have all the cards.
So, we will take those cards in our sample space.
But, the condition we are given here is that there should
be a king.
So, we will draw the probability of drawing from the king.
That is why we have taken 4C2.
Probability of selecting two kings out of four kings.
This will be your Ne.
So, what is your probability?
Probability of getting both kings.
So, this will be 4C2 divided by 52C2.
If you calculate this, it will be 4 factorial divided by 2
factorial divided by 4 minus 2.
That is 2 factorial.
52 factorial upon 50 factorial because 50 minus 2 factorial
and into 2 factorial.
So 2 factorial to 2 factorial cancel.
How can you write 52 factorial?
Here comes your 4 factorial divided by 2 factorial upon.
You can write 52 factorial as 52 into 51 into 50 factorial
because below you have 50 factorial.
So 50 to 50 will cancel.
What is left with you? 52 into 51.
And above 2 to 1 will cancel 2.
So 4 into 3 that would be 12.
And 52 into 51.
You can divide it.
Divide it by 3.
4 times 3 is 1 and 21 is 7.
Now 4 times 1 is 4 and 4 times 1 is 12.
So 1 upon 13 into 7.
What will be the final answer?
Both the cards being king.
Both the cards we have selected are kings.
So when we have a selection case and we have to make a lot
of samples.
Then we use the combination.
In which the order doesn't matter.
Now if you show both the cards to someone.
Then he will not know.
It doesn't matter at the last.
What do you want? Both should be kings.
Let's move on to the next question.
Let's read the question once.
When 3 fair coins are tossed together.
How many fair coins? 3 coins.
As soon as 3 coins come, you can take out the elements of
the sample space.
What will be the total number? 3 fair coins.
So 2 power n that is 2 power 3.
Which is 8.
So how many elements will come in the sample space? 8
elements.
We will write those 8 elements once.
Because it is 8 and not much.
So we will write it once.
So what will come?
First we will keep h constant.
So in second h and in third t.
Then h, h and in third also h.
So first we kept these two.
Now h, t, h can come.
We kept these two.
Then next we will keep t, t, h.
We are replacing h with t.
So t, h, t can come.
So t, t, t can come.
Then what are the remaining elements? h, t, t can come.
Let's see which case is left.
h here and double t here.
Then t here and double h here.
How many will be there? 6 and 2, 8.
You will see that there are no more cases.
These are the cases.
This is total 8.
What is the probability of getting at least two tails?
What you have to find out is at least.
Here you have two words in probability.
That is one is at least and other is at most.
Now what is the difference between these two?
If we say that at least 5 marks should come.
At least 5 marks should come out of 10.
So 5 marks should come.
Or 6 marks should come.
If it is bigger than 5, then you will take that case also.
Then 7, 8, 9, 10 can also come.
So in the case of at least, whatever number is given to
you, you will take all the numbers in front of it.
There should be at least 5 marks.
If there are more than 5 marks, then no problem.
Now if I do the same thing that at most 5 marks should come.
At most means more than more.
So more than more 5 marks should come.
That is 0 marks can also come.
So in that you will do counting from 5 to below it.
So 5 marks can also come.
4 marks can also come.
That is also valid in this.
So 3 marks can also come.
2 marks can also come.
1 mark can also come.
And 0 marks can also come.
If 5 marks are given more than more, then more than more.
Less can also come.
So less will come 4, 3, 2, 1, 0.
We will take these cases in this.
So I hope you have cleared the difference between at least
and at most.
Now what is the probability of getting at least 2 tails?
You should have at least 2 tails.
So at least 2 tails should be there.
This case will also work.
And above 2 tails will also work.
That is 3 tails.
You can go up to any number above 2.
But you can go up to maximum 3.
Because you have only 3 numbers.
You are flipping only 3 coins.
You are not tossing more.
So either you have 2 tails or more than 2.
That is 3 tails.
So you have to find out so many cases in this.
So what is it?
If there are 2 tails in this, then 1 case is done.
If there are 3 tails in this, then 3 cases are done.
If there are 2 tails in this, then 3 cases are done.
If there are only 1 tail in this or no tail in this.
There is no tail in this.
So we have at least 2 cases.
In which this came, this came and this element came.
And there is only 1 case with 3 tails.
That is 3TTT.
So this came.
So in total how many elements did we get in this event?
That is TTH, THT, TTT and HTT.
We got these.
So what is NE? 4 elements.
So what is the probability?
NE by NS.
That is 4 divided by 8.
Which is 4 by 8, 1 by 2.
So you will get the final answer.
So what is the probability? 1 by 2.
That is 1 by 2.
This was an easy question.
In this we just had to tell you the meaning of at least.
At least and at most.
When these words come.
So what do we do?
If the same question, at least you have done.
At least 2 tails.
If I change the same question a little.
I ask at most 2 tails.
What am I asking you?
At most.
Let me rub this.
I am asking you at most 2 tails should come.
At most 2 tails will give you.
One is 2.
At most means maximum 2 tails.
So either 2 tails come.
Or 1 tail comes.
Or 0.
You will go up to 0 tails.
So you will go up to 1 tail.
So you will go up to 0.
Which has no tail.
Maximum 2.
So maximum 0, 1, 2.
So we will not take any number greater than 2.
At most is asked.
So 0, 1, 2 case will come.
Now if we make it.
What will be the number of elements of it?
In this event.
Let's give the name of this event as F.
Because E is the name of this event.
So NF.
Which elements will come in this?
At most 2 tails.
So first we will see 0 tail.
In 0 tail you will have this element.
HHH.
Because it has no tail.
Then you have 1 tail case.
Which has 1 tail.
That is HHT and HTH.
Right? And someone else has 1 tail.
Obviously, this also has 1 tail.
THH.
So 3 will be this.
So you will have 1 element.
That is HHH.
In this you will have 3 elements.
So I will not write the name.
I will write it directly.
That there are 3 elements.
And in 2 tails.
In 2 tails we saw that we got 3 elements.
1 this, 1 this and 1 this.
You will have 3 elements of this also.
So in total how many elements you will have? 3 plus 3 is 6.
7 by 8 7 by 8
At most two tails
At most two tails 3 tail option 7 by 8
I hope you got the difference
Moving on to next question
Out of 17 candidates, how many are total candidates?
Total candidates 17 Out of 17, there are 8 boys and 9 girls
2 persons are to be selected for the job 2 persons 2
persons are to be selected for the job 2 persons
Out of 17, you need 2 people
You have to select 2 people among 17 candidates Find the
probability that at least one of the selected person will
be a girl
What is the event?
At least one of the selected person is a girl
What is the event?
What are the elements in this event?
At least one girl
Out of 17, we need 2 candidates Out of 17, at least one of
the candidates should be a girl Both of them can be a girl
What is the case?
Either one girl or one boy Second case is
You have one girl and another girl
That is girl-girl case
This was the case of girl-boy
One girl and one boy
How can you arrange this?
You can write girl-boy or first boy and then girl
But if we don't need to mention the order
We can use the combination formula
What is the event?
If we need one girl and one boy
We have to select one girl
Out of 9, one girl can be selected
The new rule is
Probability will be same as combination
We have studied addition and multiplication principle
If both the cases are same
One boy and one girl
One girl and one boy
In the case of AND, it will be into
In the case of OR, it will be plus 9C1 into 8C1 Divided by
We will divide now
We will not divide the probability now
This is the number of possibilities
In how many ways it can be
Similarly, in how many ways it can be
That is, you have to select 2 out of 9
You have to select 2 out of 9
Now, what will you do with these two possibilities?
You will add because there is sign of OR
So, you will add these two
So, this is NE
Means, how many possibilities can be in E?
In events, number of favorable outcomes
It will come in that
Number of favorable outcomes
Now, you should get total number of possible outcomes
That will be, you have 17 total candidates How many you
have to choose from that?
You have to choose 2
So, what will be NS? 17C2 How many total candidates you
have to select?
You have to select 2 out of them
So, what will be the probability?
Probability of E
9C1 x 8C1 9C2 Whole divided by 17C2
I hope you will solve this Otherwise, 9C1 x 8C1
How many you will get? 9 x 8
Because, whenever you will do this 9C1 will be 9! 9-1 8! 1!
So, always 9NC1
Or RC1
What will come?
It will be equal to R
Or whatever is the number above
So, it will cancel out 8! And only 9 will be left
So, plus 9! 7! 9! 7! into 2! divided by 17! upon
15! into 2! If you solve this 9 x 8 plus
After this, 7! 9 x 8 into 7! 7! 7! Cancel and in the
divide, how can we write 17 factorial, 17 x 16 x 15
factorial, 15 factorial cancels 15 factorial, 2 and here 8,
so 17 x 8, ok.
Now what you do, solve it, divide it by 4, 9 x 2, that is 9
x 2, 18, plus 9, upon me, 17 x 2, 14, 1, 2 under 2, 3.
So this is your 16 x 17, 1, 27 x 34.
This is your final answer.
I hope it will be clear to you, how we took the
combination, how we took the event, then divided by 17 x 2.
Next question, what is the probability of getting a sum 9
from 2 throws of a die.
Whose case is this, do you remember? I had done 2 throws,
in whose sample space, how many elements were there, 6 x 6,
that is 36, 36 elements were there in the sample space, so
we have to do the probability that what will be the sum,
sum of those 2 numbers
should come, you have 9.
Ok, so if you make a sample space, so if you see in it, who
can be the sum of 9, one if you have
9, so in the next one, 8 should come, so that your sum
becomes 9, but it is not possible, so 1 will not be with
anyone.
Now see 2, what will you add in 2, 9 will come, 7 will be
added, so 7 will not come in the dice, so 2 will not be
with anyone, next 3, what will you add in 3, 9 will come,
if you add 6 in 3, 9 will come, so who can be the pair of
3, 6 can be made with
6, similarly 6 can also be made with 3, so 2 pairs are
there, 3, 6 and 6, 3.
Then 4, what will you add in 4, if you have 9, you will add
5 in 4, so one more pair is there, 4, 5 or 5, 4.
Ok, after 4, 5 is there, if you add 4 in 5, then obviously
I have already written that, so if you add 6 in 6, then 3
will come, so how many total elements are coming in the
event, this is your E, E and E, how many you have, 4
elements, so what will you do, divide it by NS, so 4
divided by 36, how many will come,
1 by 9.
I hope this can be used for anyone, if you are asked about
sum 9, similarly you can ask about sum 12, 17, you can ask
about any sum, not 17, maximum sum will be 6 or 6, 12, so
you can ask anything up to sum 12, and what can you ask,
you can ask about
prime, even
or odd, there will be such cases, but what you have to keep
in mind, you should keep in mind sample space, what will be
there, in any question you should know two things very
well, to find sample space and what elements will come in
event, means what
will be your favourable outcomes, you should know that.
Next question, a bag contains 4 white, 5 red and 6 blue
balls, so a bag has 4 white balls.
This question will look like a permutation and combination.
So we will do permutation and combination only.
We will find out the probability at the end.
So 4 white balls, 5 red balls and 6 blue balls.
3 balls are drawn at random from the bag.
Total we have 15 balls.
You have to draw 3 balls from these 15 balls.
Find the probability that none of them is red.
None of these balls should be red.
That means you have to draw 3 balls from either of these 3
balls.
Now, there are 3 balls.
You can fill 3 balls in any way.
But none of these balls should be red.
You can also do it in a different way.
I will leave it to you.
You can do it in a different way.
Forget about the red one.
You have to draw 3 balls from either of these 3 balls.
So you will have different cases.
You can take 3 balls from 4 or 3 balls from 6.
You can take 1 ball from 4 or 2 balls from 6.
You can do it in this way.
You can multiply all these cases and add them.
So you will get the result.
Do it from that case.
I will tell you the short method from that case.
First, look at the sample space.
How will we find the sample space?
There are 15 balls in total.
We have to select 3 from 15.
So, we will get 2 from 15.
Next, find the probability that none of them is red.
So, we did a case of complement.
None of them is red.
If we find the case that all the balls are red.
If we minus the red case.
Whose case? Red case.
If at least one ball is red.
At least one ball is red.
At least one red case.
If we minus at least one red case.
Because at least one red has 1 red, 2 red and 3 red.
If we minus that, whose case will we get?
None of them is red.
If we minus at least one red case.
None of them is red.
So, we have to find at least one red.
Because we know the probability of total.
Probability of total is 1.
100% sure that is 1.
So, if we minus at least one red from total.
We get at least one red.
For at least one red, if there is one red.
Or there are two reds.
Or there are three reds.
For at least one red, there will be three cases.
One red, two reds or three reds.
If we need three balls, then we get three reds.
If we need one more ball from two reds.
That one ball can be of any of these two cases.
Either it can be white or it can be red.
It can be white or it can be blue.
One ball.
Now, if there is one red.
Then, either the other two can be white or blue.
One white, one black, we don't care about that.
We just write this.
Now, how will we find out their particular? Now, see.
One red.
I will rub this.
I will write this later.
First, I will tell you this.
How will we find out their particular? See.
One red.
When there is one red ball.
Then, we are choosing one out of five.
Then, what will happen? 5C1.
We will multiply it.
Because, it is running simultaneously.
And, 2 white or blue.
Now, we will multiply 2 white or blue.
We will multiply white and blue.
That is 10.
Now, any two balls from those 10 will work for us.
They can be red or blue.
One red, one blue.
One blue, one red.
It can be anything.
So, for that, we will multiply it by 10C2.
Did you understand? What is 10? 10 is the addition of 4
white and 4 blue.
And, 2.
Because, we have chosen one ball.
In total, we have to select three balls.
So, what did we do?
We selected 2 out of white and blue.
Did you understand?
Now, see.
The addition below is equal to 3.
And, the sum above is equal to 5 or 10.
It is equal to the total number of balls.
Next case.
2 red.
So, we need 2 reds.
5.
So, 5C2 will be from red.
We need 2 from red.
And, we need one ball.
White or blue.
It can be anything.
So, we selected one ball from the remaining 10.
This is the case.
Both are multiplied.
Then, 3 red.
So, we are subtracting 5 from red.
We don't have to take the rest.
Because, we only need red.
So, 5C3.
Now, these are the cases.
What are the other cases?
Either 1 red, or 2 red, or 3 red.
At least.
So, what will we do with all these?
We will add them.
If we add all these, we will get the probability.
That we will get at least 1 red ball.
At least.
Whose probability are we getting?
At least 1 red.
Okay? If we add all these.
Then, we will get Ne.
Ne.
Not Ne.
Sorry.
We will subtract Ne.
None of the red.
It means, we will get the complement.
We will get the Ne complement.
Ne complement.
Because, what is E?
None of them is red.
So, we will subtract 1 from all these.
Okay? What will we get as the probability? Probability of
EC.
How much will you get?
You will sum all these.
You will divide it by 15C3.
Okay? You will calculate all these.
You will get 5C1.
That is 5.
Into 10C2.
If you subtract 10C2.
10 factorial divided by 8 factorial.
Into 2 factorial.
8 will be subtracted from here.
2 will be subtracted from here.
So, you will get 45.
Into 45.
Plus 5C2.
5 factorial divided by 3 factorial.
2 factorial.
3 factorial will be subtracted from here.
Then 2.
You will get 10.
If 10C1 is not 10.
Then 10 into 10 will be 100.
Plus 5C3.
5C3 will be 5 factorial.
3 factorial divided by 2 factorial.
That will be subtracted.
And this.
You will get 10.
So, you have to solve this.
What will you get? 45 into 5.
That is 25.
20.
225.
Plus 110.
Divided by 15C3.
15C3.
15 into 14.
Into 13.
Divided by 3 factorial.
That is 3 factorial.
3 into 2.
Solve this.
This is cancelled by 5.
This is cancelled by 7.
5 into 7 into 13.
Solve this.
Add this.
You will get this.
This is the probability.
You have to do 1 minus this.
So, 1 minus this.
If you complete this.
You will get this answer.
Solve this.
Calculation will be like this.
35 into.
I will erase this.
35 into 13.
15.
9, 1, 9, 10.
Into 35.
5 5 4 55
Divide by 455 and what do you have?
5 3 and 3 335 by 455
Okay, you have to do this in short.
Cancel out 1 minus
If you do, then your total probability will come.
This will be your final answer. 1 minus 335 divided by 455.
Okay, from 5, you can do it in simplest form.
So, these were the questions of today's probability.
With these, we were able to learn every new concept that
has been asked.
So, we will meet you in the next video. Till then, you guys
keep on practicing as much as you can.
A lot of topics have been covered. I hope you will keep on
practicing them side by side.
So, see you in the next video. Thank you.
I'll make sure to spread the word to all my friends.