This should be a select one multiple choice question.
There are altogether 10 integers namely disk labelled 1 to 10 inclusive therefore total number of outcomes for choosing 4 integers out of 10 is \(10C4 = \frac{10*9*8*7}{4*3*2} = 210\)
Now lets look at the ways in which we can chose
4 integers such that the range becomes
7.
There are only
3 possibilities how a range of
7 could occur when
4 numbers are chosen
1st. \((1,8)\): If among the 4 number chosen 2 of the numbers are 1 and 8 and the rest of the 2 numbers are from {2,3,4,5,6,7}. Notice that 9 cannot be chosen because it results in a range of 8.
2nd. \((2,9)\): Rest of the 2 numbers should be from the set {3,4,5,6,7,8}
3rd \((3,10)\): Rest of the 2 numbers should be from the set {4,5,6,7,8,9}
Now let us take the case for any 1 of 3 possible option.
(2,9)
There are 4 slots to chose the number. First 2 should be (2,9)
Hence 2c2 = 1
Next 2 numbers should be from the set of remaining 6 integers {3,4,5,6,7,8}
Hence \(6c2 = 15\)
Therefore for 1 of the possibility the required outcome = 15 ways
There are 3 possibilities therefore total possible ways = \(15*3 = 45\)
Therefore the probability is \(\frac{45}{210}\) which reduces to \(\frac{3}{14}\)
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