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23 May 2019, 01:53
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59% (02:34) correct
41%
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The only contents of a container are 10 disks that are each numbered with a different positive integer from 1 through 10, inclusive. If 4 disks are to be selected one after the other, with each disk selected at random and without replacement, what is the probability that the range of the numbers on the disks selected is 7 ?
A) \(\frac{1}{7}\)
B) \(\frac{3}{14}\)
C) \(\frac{2}{7}\)
D) \(\frac{1}{2}\)
E) \(\frac{15}{28}\)
A) \(\frac{1}{7}\)
B) \(\frac{3}{14}\)
C) \(\frac{2}{7}\)
D) \(\frac{1}{2}\)
E) \(\frac{15}{28}\)
23 May 2019, 03:09
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We have ten disks numbered 1 to 10 inclusive. First of all the total possibilities (events) is 10C4 = 210. Now in order to get a range of 7, the possible upper and lower limits are (1,8), (2,9) and (3,10).
Since order is not important, I will focus on one of the sets and multiply the possibilities by 3. So let’s take (1,8). Other possible disks that can be drawn to satisfy this range are: 2,3,4,5,6,7. Remember we already made two selections i.e. 1,8. So the total events is 6C2. (There are 6 other disks that can be combined in 2 ways). 6C2 is 15, 15x3 is 45.
Hence answer is 45/210 =3/14.
Posted from my mobile device
Since order is not important, I will focus on one of the sets and multiply the possibilities by 3. So let’s take (1,8). Other possible disks that can be drawn to satisfy this range are: 2,3,4,5,6,7. Remember we already made two selections i.e. 1,8. So the total events is 6C2. (There are 6 other disks that can be combined in 2 ways). 6C2 is 15, 15x3 is 45.
Hence answer is 45/210 =3/14.
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06 May 2024, 21:06
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