00:01
Hi, in this question, given that number of subject reported from drug a equals 20 and number of drugs reported by drug b equals 20 and number of subject reported relief from drug c equals 30 and n of a intersection b equals 7 and n of a intersection c equals 13 and n of b intersection c equals 14 and n of a intersection union b union c equals 42.
00:41
So in part a, we have to find the number of people got relief from none of the drugs.
00:48
So which can be written as 50 minus 42, which is equal to 8.
01:04
Next move on to part b.
01:06
Here number of people got relief from all the three drugs, which is n of a union b union c equals n of a plus n of b plus n of c minus n of a intersection b minus n of a intersection c minus n of b intersection c plus n of a intersection b intersection c.
01:32
So here we have to find the value of n of a intersection b intersection c.
01:37
On substituting all the known values in this, then we get 42 equals 20 plus 20 plus 30 minus 7 minus 13 minus 14 plus n of a intersection b intersection c.
01:52
Hence conclude that n of a intersection b intersection c equals 42 minus 36, which is equal to 6...