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Three numbers are chosen at random without replacement from 1 to 10. The probability that the minimum of the chosen number is 3, and their maximum is 7, is?

Three numbers are chosen at random without replacement from 1 to 10. The probability that the minimum of the chosen number is 3, and their maximum is 7, is?

Grade:12th pass

4 Answers

Jayant Kishore
144 Points
8 years ago
Number of elements in sample space = 10C3 =120 
According to the question 3 & 7 must get chosen and the third no. should be between them (ie 4 or5 or 6)as per condition .
So, no of favourable cases =  3C1 =3
Therefore required probablity = 3/120 = 1/40
Approve the answer if correct .
 
Jayant Kishore
144 Points
8 years ago
Alternatively , you can proceed this way.
No. of elements in sample space = 10 x 9 x 8 for selection of 3 numbers one by one without replacement.
According to the question 3 & 7 must get chosen and the third no. should be between them (ie 4 or5 or 6) as per condition .
So , favourable selection may be as 3  , 7 , X
where X denotes either of 4 or 5 or 6. This can be done in 3 ways.
But , here since we are selecting one by one , IN THIS CASE ORDER OF SELECTION MATTERS
So, no of favourable cases = 3! x 3
Therefore required probablity = 3! x 3/ (10 x 9 x 8) = 1/40.
NOTICE THE DIFFERENCE . 
 
 
Humanshu Parkhade
36 Points
7 years ago
P(min 3 union max 7)= p(min 3)+ p(max 7) - p( min3 intersection max7). Now put the values........... 8C3 by 10C3. +7C3 by 10C3 -5C3 by 10C3 .... Solve this u get answer correctlt
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the attached answer to your question.
Number of ways in which 3 random numbers can be drawn from 1 to 10 = 10C3 = 120 
It is given that 3 & 7 must get chosen and the third no. should lie between them
Therefore, the possible third number can be either of 4, 5 or 6
So, favourable cases =  3C1 = 3
Therefore required probablity = 3/120 = 1/40
 
Hope it helps.
Thanks and regards,
Kushagra

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