Probability

Here, total possibilities, S = {1, 2, 3, ...,15} or n(S) = 15

Let E = event of getting a multiple of 2 or 3 = {2,3,4, 6 ,8, 9,10, 12,14, 15}or n(E) = 10.

Then, the required probability, P(E) = n(E)/n(S) = 10/15 = 2/3.

2 crayons can be drawn out of 15 in ¹⁵C₂ = 105 ways. 
One green crayon and one yellow crayon separately can be drawn in ⁷C₁ = 7 and ³C₁ = 3 ways, respectively.
Since they are exclusive and independent events, taking one green and one yellow crayon together will make the favourable number of ways equal to ⁷C₁ x ³C₁ = 21 ways.
Hence, required probability, P = ⁷C₁ x ³C₁/(¹⁵C₂) = 21/105 = 1/5

Total number of marbles = (3 + 2 + 2) = 7
Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 marbles out of 7 = ⁷C₂ = (7 x 6)/(2 x 1) = 21
Let E be the event of drawing 2 marbles, none of which is blue (i.e. either yellow or green marble is drawn).

Therefore, n(E) = Number of ways of drawing 2 marbles out of (2 green + 3 yellow) marbles = ⁵C₂ = (5 x 4)/(2 x 1) = 10
P(E) = n(E)/n(S) = 10/21

Probability of getting a 1 on the first throw = 1/6
Probability of getting a 5 on the first throw = 1/6
Therefore, using Addition theorem of probability, the probability of getting number 1 or 5 on the first throw irrespective of the result of second throw = (1/6) + (1/6) = 1/3

In two rolls of a dice, total number of possibilities, i.e. n(S) = (6 x 6) = 36.
Let E be an event of getting number 1 or 5 in both the throws = {(1, 1), (1, 5), (5, 1), (5, 5)} or n(E) = 4.
Then, the required probability, P(E) = n(E)/n(S) = 4/36 = 1/9
Incorrect answer

Required probability, P(E) = n(E)/n(S) =120/231 = 40/77

The probability of the first drawn cube being green are 3/5.
As the first cube is replaced back into the box, the probability of second drawn cube being yellow is 2/5.
Since both the events are independent of each other, the combined probability, using Multiplication theorem of probability is (3/5)x(2/5) = 6/25.

If all the three cards are considered, then originally they had 4 Hs and 2 Ts.
Since one H has already been revealed, now the probability of getting one more H out of 3 Hs and 2 Ts is to be found OR (3 Hs)/(3 Hs + 2 Ts) = 3/5

Let S be the sample space, i.e. possible outcomes of four independent tosses of an unbiased coin.
Then, n(S) = 2 x 2 x 2 x 2 = 16
Out of these 16 outcomes, there is only one possible outcome where one gets all the four tails.
In all the remaining (16 - 1 = 15) cases, there will be at least one head.
Therefore,
Required probability = 15/16

Let S be the sample space and E be the event of selecting 3 cards of diamonds.
Number of ways of drawing 3 cards out of 52 = n(S) = ⁵²C₃
Number of ways of drawing 3 cards of diamonds out of 13 = n(E) = ¹³C₃
Then, required probability P(E) = n(E)/n(S) = ¹³C₃ / ⁵²C₃ = 11/850

If all the 83 balls are considered, then originally there were 80 blue and 3 yellow balls.
Since one of the blue ball has already been revealed, now the probability of getting one yellow ball out of 3 yellow and remaining (80 - 1) or 79 blue balls = 3/(3+79) = 3/82

Number of ways of drawing 3 tickets out of 50 = ⁵⁰C₃
Number of two-digit tickets = 50 - 9 = 41 (since number of one-digit tickets, i.e. tickets marked 1 to 9 = 9)
Number of ways of drawing 3 tickets out of 41 = ⁴¹C₃
Therefore, probability of getting two-digit numbered ticket = (⁴¹C₃)/(⁵⁰C₃)
Probability of getting at least 1 one-digit numbered ticket = 1 - probability of getting two-digit numbered ticket
So, required probability = 1 - (⁴¹C₃)/(⁵⁰C₃)

As the cards are drawn simultaneously, there is no replacement made.
Let S be the sample space and E1 be the event of selecting 2 hearts and E2 be the event of selecting 3 black cards.
Number of ways of drawing 5 cards out of 52= n(S) = ⁵²C₅ Number of ways of drawing 2 hearts out of 13 = n(E1) = ¹³C₂ Number of ways of drawing 3 black out of 26 = n(E2) = ²⁶C₃
Since both the events are independent of each other, then the probability of drawing 2 hearts and 3 black cards = P(E) = n(E1)x n(E2)/n(S) = (¹³C₂)x(²⁶C₃) / (⁵²C₅ )

Let S be the sample space.
Then, n(S) = Total number of candies = (30 + 20 + 16) = 66
Let E be the event that the candy drawn is neither strawberry nor vanilla flavoured = Event that the candy drawn is chocolate flavoured.
Then, n(E) = Number of chocolate flavored candies = 16
P(E) = n(E)/n(S) = 16/66 = 8/33

Let the order of correct sizes be (1, 2, 3).
The vests can be delivered to the customers in the following ways:
(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) and (3, 2,1) = 6 ways.
The cases (1, 2, 3), (1, 3, 2), (2, 1, 3) and (3, 2, 1) correspond to at least one right sized delivery OR 4 ways. Therefore, the required probability = 4/6 = 2/3