A bag contains 14 identical balls of which 5 are red, 3 black and 6 white balls. Four balls are drawn at random from the bag. Find the probability that (i) one is white
Out of "14" balls "4" balls can be drawn in "^{14}C_4=1001" ways.
Now one white ball can be drawn in such ways
(1) "1" white "3" red balls
(2) "1" white "3" black balls
(3) "1" white "1" red "2" black balls
(4) "1" white "2" red "1" black balls
Therefore, "1" white "3" red balls can be drawn in "( ^6C_1\u00d7 ^{5}C_3)""=10" ways.
"1" white "3" black balls can be drawn in "( ^6C_1\u00d7 ^{3}C_3)""=6" ways.
"1" white "1" red "2" black balls can be drawn in "(^6C_1\u00d7 ^{5}C_1\u00d7 ^{3}C_2)"= "90" ways.
"1" white "2" red "1" black balls can be drawn in "(^6C_1\u00d7 ^{5}C_2\u00d7 ^{3}C_1)" "=180" ways.
Let "A" be an event such that out of four drawn balls one ball is white.
Then number of elements in favour "A" is "=(10+6+90+180)=286"
Therefore, "P(A)=\\frac{n(A)}{n(S)}=\\frac{286}{1001}=\\frac{2}{7}"
Comments
Leave a comment