Probability and Statistics L :5 | Engineering Mathematics | End GAME of Mathematics | Saurabh Thakur
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- Опубліковано 7 лют 2025
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#Probability_and_Statistics #General_Aptitude
Student dharma
WHEN REPETITION IS ALLOWED
1/2/3/4/5/6 SO SAMPLE SPACE=6^6
1.For No. to be divisible by 2 last two digits should be divisible by 2 so 2/4/6 are no.
P[no/2]=6^5*3/6^6=1/2
2. For No. to be divisible by 2 last two digits should be divisible by 4so 12 16 24 32 36 44 52 56 64 are no. So 9 Such cases
P[NO/4]=6^4*9/6^6=1/4
WHEN REPETATION IS NOT ALLOWED
1/2/3/4/5/6 SO SAMPLE SPACE=6!
1.For No. to be divisible by 2 last two digits should be divisible by 2 so 2/4/6 are no.
P[No/2]=5!*3/6!=1/2
2. For No. to be divisible by 2 last two digits should be divisible by 4so 12 16 24 32 36 52 56 64 are no. So 8 Such cases
P[N/4]=4!*8/6!=4/15
Thankyou sir
ANSWER
Given Numbers are, 1, 2, 3, 4, 5, 6
A) Repetition Allowed:
i) The Number is divisible by 2:
We have six numbers and we can repeat them so our sample space is => 6 ^ 6.
Now for divisible by 2 the last digit has to be divide by 2
_____ _____ _____ _____ _____ -----(Last Possible digit divisible by 2 are => 2, 4, 6)
6 x 6 x 6 x 6 x 6 x 3
Here Repetition is allowed so we have six choice for first 5 spaces and for the last one only three choice => (6 ^ 5) * 3
Therefore the Probability of getting a number divisible by 2 is => ((6 ^ 5) * 3)/(6 ^ 6) = 1/2
ii) The Number is divisible by 4:
We have six numbers and we can repeat them so our sample space is => 6 ^ 6.
Now for divisible by 4 the last two digits have to be divisible by 4
_____ _____ _____ _____ -------- --------(Last Possible digit divisible by 4 are => 12, 16, 24, 32, 36, 44(we can repeat the numbers), 52, 56)
6 x 6 x 6 x 6 x 8
Here Repetition is allowed so we have six choice for first 4 spaces and for the last two we have 8 choice => (6 ^ 4) * 8
Therefore the Probability of getting a number divisible by 2 is => ((6 ^ 4) * 8)/(6 ^ 6) = 1/2
B) Repetition NOT Allowed:
i) The Number is divisible by 2:
We have six numbers and we can't repeat them so our sample space is => Permutation of 6 numbers => 6! = 720.
Now for divisible by 2 the last digit has to be divide by 2
_____ _____ _____ _____ _____ -----(Last Possible digit divisible by 2 are => 2, 4, 6)
5C1 x 4C1 x 3C1 x 2C1 x 1C1 x 3
=> 5 x 4 x 3 x 2 x 1 x 3
Here Repetition is not allowed so we have five choice for first 1 one, four choice for first 2 one, Three choice for 3rd one, Two choice for 4th one & only One choice for 5th one and for last
one only three choice => 5 * 4 * 3 * 2 * 1 * 3 = 360
Therefore the Probability of getting a number divisible by 2 is => 360C1/ 720C1 = 1/2
ii) The Number is divisible by 4:
We have six numbers and we can't repeat them so our sample space is => Permutation of 6 numbers => 6! = 720.
Now for divisible by 4 the last two digits have to be divisible by 4
_____ _____ _____ _____ -------- --------(Last Possible digit divisible by 4 are => 12, 16, 24, 32, 36, 52, 56)[44 will not occure because we can't repeat the numbers)]
4C1 x 3C1 x 2C1 x 1C1 x 7
=> 4 x 3 x 2 x 1 x 7
Here Repetition is not allowed so we have four choice for first 1 one, Three choice for 2nd one, Two choice for 3rd one & only One choice for 4th one and for last two we have 7 choices
=> 4 * 3 * 2 * 1 * 7 = 168
Therefore the Probability of getting a number divisible by 2 is => 168C1 / 720C1 = 7/30
Nice explanation, but for divisibility by 4 in both cases (that is when repetition allowed and not allowed) why you have not considered 64. The answer will be 1/4 and 4/15 respectively. Kindly check.
Superb explained except 1 mistake👍👍
@@jayasimhack2410 yup you are right
35:14 Big fan of your singing too from now😁..
where can i get the notes sir for this topics
Great
When repeatation allowed
Total sample = 6 != 720
Number formed in last two digits divisible by 2 are 12,14,16,24,26,32,34,36,42,46,52,54 ,56, 62,64
P(E)= (4!×15)/6! = 1/2
Divisible by 4
12 ,16,24 32,36,52,56,64
P(E)= (4!×8)/6!= 4/15
When repetation allowed
Number formed in last two digits divisible by 2 are
12,14,16,22, 24,26,32,34,36,42,44, 46,52,54 ,56, 62,64 ,66
P(E)= (6×6×6×6×18)/6×6×6×6×6×6 =1/2
DIVISIBLE BY 4
12 ,16,24 32,36, 44,52,56,64
P(E)=(6×6×6×6×9)/6×6×6×6×6×6 = 1/4
For numbers which are divisible by 2 we have to consider only the last digit but why you have considered last two digits?
1. For not repetition allowed
U can consider the digit last will be same like this
5! For starting 5 digit and for last digit have possibility 2,4,6 make divisible by 2. So we say that P(%2) = (5! * 3)/6! = 1/2
Choices are at last 2 digit is 24, 36, 12, 64, 56, 16, 32, 52
P(%4) = (6^4 * 8)/6 ^6 = 4/15
Sir, I tried doing a question applying the same approach that I have learned so far from your videos but it seems doubtful. Can anyone try the qsn '2 dice are tossed 10 times. What is the probability that pair (i, i) is obtained at least 1?' ( avoid binomial approach) . Use the basic probability method that we learned in this playlist.
Sir Dil kush ho gya
Lectures Starts at 1:35
24:14, sir you have forgotten "48"
8 hai hi ni
Sir aap padhane se jyda overacting kyu karte hai