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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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