Probability Part 3

Probability Part 3

Ex. A card is drawn at random from a pack of 52 cards, What is the probability that it is (a) A card of Red Suit ? Solution : There are 26 cards of Red Suit (b) An honour card of Black suit ? Solution : There are 16 honour cards out of which 8 are of Black suit and 8 are of Red Suit. So n(E) = 8 , n(S)=52 (c) A card is drawn and its number is multiple of 2 Solution : E = 4 (2)'s + 4 (4)'s + 4 (6)'s + 4 (8)'s + 4 (10)'s So, n(E) = 20, n (S) = 52 (d) A king or a queen ? Solution : There are 4 kings and 4 Queens in 52 cards (e) A king of black suit ? Solution : There are 2 kings in black suit ( King of Spade and King of Club ) Ex. A bag contains 4 red, 3 yellow and 5 green balls. 3 balls are drawn randomly. What is the probability that balls drawn contain (a) Balls of different colors ? Solution : Toatal numbers of balls = 12 (b) Exactly two Red Balls ? Solution : Here only three balls are to be drawn out of which condition is of Exactly two Red balls, (c) No Red balls ? Solution : Now three balls can be selected from 3 Y + 5 G balls Ex. A bag contains 4 Red balls and 5 Green balls. Two balls are drawn at random. Find the probability that they are of the same colour ? Solution : Let S be the sample space and E be the event, so n(E) = ( Number of ways of drawing 2 balls of Red ) OR ( Number of ways of drawing 2 balls of Green ) Ex. A three-digit number is formed with the digits 1, 2, 3, 4, 5 at random. What is probability that number formed is (a) Divisible by 2 Solution : From the given digits 1, 2, 3, 4, 5 numbers formed is : For divisibility with 2, even number or 0 should appear at unit place, here 2, 4 are even numbers and can occupy unit place in 2 ! ways, Rest 2 place can be filled in : Ex. Not divisible by 2 ? Solution : P (Not divisible by 2 ) = 1 - P (Divisible by 2 ) Ex. Divisible by 5 ? Solution : A number ends with 5, 0 then the number will be divisible by 5 Here only 5 is present, end place will be fixed by 5 so, Ex. The letters of the word CASTIGATION is arranged in different ways randomly. What is the chance that vowels occupy the even places ? Solution : Vowels are A I A I O,  C   A   S   T   I   G   A   T   I   O   N (O)(E)(O)(E)(O)(E)(O)(E)(O)(E)(O) So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :