We know that in English alphabet, there are 26 letters (5 vowels + 21 consonants).
So, total number of outcomes = 26
Let E be the event of choosing a consonant.
\(\therefore\)  Number of outcomes favourable to E = 21
Hence, required probability = \(P(E)=\frac{21}{26}\)

Total number of tickets = 40 and multiple of 5 are 5, 10, 15, 20, 25, 30, 35, 40
P(a number which is a multiple of 5)
\(=\frac{8}{40}=\frac{1}{5}\)

Number of ways to draw a card by 1st player = 1000
perfect square greater than 500 are 529, 576, 625, 676, 729, 784, 841, 900, 961
\(\therefore \)Probability of 1st player winning the prize = \(\frac { 9 }{ 1000 } \)  
when 1st has won the prize then cards left = 999 (because card is not replaced)
Cards with no. which a perfect square greater than 500 = 8 (l winning number removed)
\(\therefore \)Probability of second player wins a prize when the first player has won =\(\frac { 8 }{ 999 } \)

Total number of enevelopes = 1000
Let A = envelope contains no cash
Number of envelopes containing no cash
 = 1000 - (10 + 100 + 200) = 690
\(\therefore\) P(A) = \(\frac{690}{100}=\frac{69}{100}=0.69\)

Yes, P (even number) = \(\frac{50}{100}=\frac{1}{2};\)
   P (odd number) = \(\frac{50}{100}=\frac{1}{2}\)

Total cases are 8. (GGG), (GGB), (GBG), (GBB), (BGG), (BGB), (BGG), (BBB)
no girl = \(\frac{1}{8}\); one girl = \(\frac{3}{8}\);
2 girls = \(\frac{3}{8}\); 3 girls = \(\frac{1}{8}\)

For Apoorv if he getting (6, 6) then he gets = Number 36
Then Probability of getting (6, 6) = \(\frac{1}{36}\)
But Peehu got 36 when he got 6, 
Probability of getting 6 = \(\frac{1}{6}\)

When two dice are thrown at the same time, then all possible outcomes are 6 x 6 i.e., 36.
Number of outcomes with difference of numbers on the two is
i.e., [(1,3), (2,4), (3,1), (3,5), (4,2), (4,6), (5,3), (6,4)] =8
Required Probability \(={8\over36}={2\over9}\)

When two dice are thrown simultaneously then sample space contains (6 X 6 = 36) outcomes.
Product of numbers is less than 9.
[(1,1), (1,2), (1,3), (1 ,4), (1,5), (1 ,6), (2,1), (2,2), (2,3), (2,4), (3, 1), (3,2), (4, 1), (4,2), (5, 1), (6,1)] i.e., 16 outcomes 
\(\therefore\) P (product < 9) =\(\frac { 16 }{ 36 } =\frac { 4 }{ 9 } \)

A poor throws two dice once.
So, total number of outcomes, n(S)=36
Number of outcomes for getting product 36,
n(E₁)=1, i.e. (6,6)
∴ Probability for Apooru\(=\frac { n(E_{ 1 }) }{ n(S) } =\frac { 1 }{ 36 } \)
Also, Peehu throws one die.
So, total number of outcomes, n(S)=6
Number of outcomes for getting 36 which is square of number 6, b(E₂)=1
So, probability for Peehu\(=\frac { n(E_{ 2 }) }{ n(S) } =\frac { 1 }{ 6 } =\frac { 6 }{ 36 } \)
Hence, Peehu has better chance of getting the number 36.