Only iced tea | T W C | – – = |T| -|T∩W| - - |T∩C| |T∩W∩C| = 66 36 36 + 24 = 18 – – The Venn diagram indicates that there are 18 passengers who drink only ice tea. So |A| 153 Therefore p (A) = |A| / |S| = 153 / 7140 = 51 / 2380. Exactly 2 of the 3 type Only T and C = - |T∩C| |T∩W∩C| ………………….i 36 24 = 12 – Only C and W = - |W∩C| |T∩W∩C| ………………..ii 36 24 = 12 – Only W and T - =|W∩T| |T∩W∩C| ………………iii 36 24 = 12 – Adding i, ii, iii we get = 12 + 12 + 12 = 36 = |B|
Only iced tea | T – W – C | = |T| -|T∩W| - |T∩C| - |T∩W∩C| = 66 – 36 – 36 + 24 = 18 The Venn diagram indicates that there are 18 passengers who drink only ice tea. So |A| 153 Therefore p (A) = |A| / |S| = 153 / 7140 = 51 / 2380. Exactly 2 of the 3 type Only T and C = |T∩C| - |T∩W∩C| ………………….i 36 – 24 = 12 Only C and W = |W∩C| - |T∩W∩C| ………………..ii 36 – 24 = 12 Only W and T =|W∩T| - |T∩W∩C| ………………iii 36 – 24 = 12 Adding i, ii, iii we get = 12 + 12 + 12 = 36 = |B|
problems-on-countable.docx