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Total numbers of numbers in the draw = 20
Numbers to be selected = 6
n(S) =
Let E be the event that six numbers match with the six numbers fixed by the lottery committee.
n(E) = 1 (Since only one prize to be won.)
Probability of winning =

Sample space when three coins are tossed: [Same as a coin tossed thrice!]
S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}
Number of possible outcomes, n(S) = 8 [Note: 2x2x2 = 8]
Let E be the event of getting exactly 2 tails = {TTH, HTT, THT}
n(E) = 3
The required probability of getting exactly 2 tails is .

Heights
Tally bars
Number
160
4
161
1
162
4
163
9
164
3
165
3
168
1
Clearly, 163 occurs 9 times. Therefore, is the mode of the data.

Let be the number of blue marbles in the jar.
Number of green marbles in the jar =
According to question,
Number of blue marbles in the jar is 8

Total number of balls in the bag = 12
Number of black balls in the bag =
According to the question,
6 more black balls are added to the bag.
Total number of balls =
And, the new number of black balls =
Also,
The required value of is 3

Let there be number of blue balls in the bag.
Number of red balls = 5
Thus, the total number of balls = total possible outcomes =
And,
According to question,
Therefore, there are 10 blue balls in the bag.

+
1
2
2
3
3
6
1
2
3
3
4
4
7
2
3
4
4
5
5
8
2
3
4
4
5
5
8
3
4
5
5
6
6
9
3
4
5
5
6
6
9
6
7
8
8
9
9
12
Total possible outcomes when two dice are thrown =
Number of times when the sum is at least 6, which means sum is greater than 5 = 15

+
1
2
2
3
3
6
1
2
3
3
4
4
7
2
3
4
4
5
5
8
2
3
4
4
5
5
8
3
4
5
5
6
6
9
3
4
5
5
6
6
9
6
7
8
8
9
9
12
Total possible outcomes when two dice are thrown =
Number of times when sum is 6 = 4

+
1
2
2
3
3
6
1
2
3
3
4
4
7
2
3
4
4
5
5
8
2
3
4
4
5
5
8
3
4
5
5
6
6
9
3
4
5
5
6
6
9
6
7
8
8
9
9
12
Total possible outcomes when two dice are thrown =
(1) Number of times when sum is even = 18

Total possible ways Shyam and Ekta can visit the shop =
(1) Case that both will visit the same day.
Shyam can go on any day between Tuesday to saturday in 5 ways.
For any day that Shyam goes, Ekta will go on a different day in ways.
Total ways that they both go in the same day =

Total possible ways Shyam and Ekta can visit the shop =
(2) The case that both will visit the shop on consecutive days.
Shyam can go on any day between Tuesday to Friday in 4 ways.
For any day that Shyam goes, Ekta will go on the next day in 1 way
Similarly, Ekta can go on any day between Tuesday to Friday in 4 ways.
And Shyam will go on the next day in 1 way.
(Note: None of the cases repeats...

Total possible ways Shyam and Ekta can visit the shop =
(1) Case that both will visit the same day.
Shyam can go on any day between Tuesday to saturday in 5 ways.
For any day that Shyam goes, Ekta will go on the same day in 1 way.
Total ways that they both go in the same day =

The possible outcomes when a die is thrown= {1,2,3,4,5,6}
Total number of possible outcomes = 6
Number of odd number, {1,3,5} = 3
And, number of even numbers {2,4,6} = 3
Hence, both these events are equally likely

The possible outcomes when two coins are tossed = {HH, HT, TH, TT}
Total number of possible outcomes = 4
Hence, the given statement is not correct. This is because one of each can occur in two different ways. Hence the mentioned events are not equally likely.

When a die is thrown twice, the possible outcomes =
Total number of possible outcomes =
The outcomes when 5 comes up at least once =
{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)}
Number of such favourable outcomes = 11
Therefore, the probability that 5 comes at least once is

When a die is thrown twice, the possible outcomes =
Total number of possible outcomes =
The outcomes when 5 comes up either on them =
{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)}
Number of such favourable outcomes = 11
Therefore, the probability that 5 will not come either time is

The possible outcomes when a coin is tossed 3 times: (Same as 3 coins tossed at once!)
{HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}
Number of total possible outcomes = 8
For Hanif to win, there are only two favourable outcomes: {HHH, TTT}
Number of favourable outcomes = 2
Therefore, the probability that Hanif will lose is

A student argues that "there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability of 1/11. We do not agree with this argument because there are a different number of possible outcomes for each sum. we can see that each sum has a different probability.

The table becomes:
Sum on two dice
2
3
4
5
6
7
8
9
10
11
12
Probability
1/36
1/18
1/12
1/9
5/36
1/6
5/36
1/9
1/12
1/18
1/36

Total number of pens = 144
Total number of defective pens = 20
She will buy if the pen is good.
Therefore, probability that she will not buy = probability that the pen is defective =

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