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A die is tossed once . Find the probability of gettingi) a number less than threeii) a prime numberiii) a number greater than two
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A die is tossed once . Find the probability of gettingi) a number less...
As question is framed, assuming a 6 sided (cube) die (with sides representing 1–6) the probability is 1 (i.e. 100% certainty).
Had the query been about two dice (pair) or just one, perhaps thrown twice, then one can calculate probability as a proportion of possible outcomes.
In my revised case, with two dice, there are 11 possible numbers (2–12) that can be thrown. They may be thrown in 36 different ways (i.e. 6x6) so 36 becomes our denominator. We now calculate the numerator or number of ways to roll a 6 or less with two dice. First we consider permutations of throwing 6 (5 ways i.e. 1&5, 2&4, 3&3, 4&2 and 5&1). Then throwing 5 (4 ways), a 4 (3 ways) etc down to a low of 2
In summary, evaluating all possible outcomes, we see that there are 15 different ways to throw a 6 or less (i.e. 5+4+3+2+1) from 36 possible outcomes. Thus probability is 15/36 or 5 in 12 = 41.666% or a little less than half a chance.
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A die is tossed once . Find the probability of gettingi) a number less than threeii) a prime numberiii) a number greater than two Related: NCERT Solutions Chapter 3 (Part - 2) - Data Handling, Maths, Class 7
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