Question

Q2] (12 Marks): Eight balls, each marked with different whole number from 2 to 9, are placed in a box. Three of balls are drawn at random (with replacement) from box. i. What is the probability that the ball with the number 5 is drawn? ii. What is the probability that the three numbers on the balls drawn are odd? What is the probability of that the sum of the three numbers on the disc is odd. iv. What is the probability of that the smallest number on the balls drawn is 5.

Answer 1

Probablility of an event happening = P(A) = (m(A))/(n(A)) , where m(A) = number ot ways event A can happen and n(A) = total number of outcomes.

The number of ways Hree balls can be drawn fa-mi 8 balles is Bez = 56 1) I balls are with out number 5 on it The ways keat 3 balls can be picked from His balls are 7ez many . Probability that no balls among the three has 5 on it 7c 3 8c3 The forobability that 5 number ball in choren in 1- 763 z 1- 8c3 7! h! 3! 8! 5!*3! 7! X5! X3! 1- 8 !*4!) 3! 1-3/8 = 33 lo

ii) A be the event of drawing three odd number balls.

The number of ways three balls can be chosen from 8 balls is given by ⁸c₃ . n(A) = ⁸c₃ .

There are four balls with odd numbers on them. i.e. balls marked with 3 , 5 , 7 , 9. 3 balls can be chosen from these four balls in ⁴c₃ ways. m(A)=⁴c₃ .

P(A) = (m(A))/(n(A)) = $\frac{^{4}c_{3}}{^{8}c_{3}}=\frac{\frac{4!}{1!3!}}{\frac{8!}{3!5!}}=\frac{4!5!}{8!}=\frac{1}{14}$

The required probability is 1/14.

ili) s odd can happen The sum of 3 numbers In hoo ways All the free neembers are odd. are even and one Two is odd. eight Noco in the Gori there are balls with u even number and 4 odd number. Chowing 3 bally cam conwist of the following four cares 1 3 balls are odd number 2 3 balls are even neember. 3) g balls are odd and one is even 4) 2 balls are even and one is odd Since there are equal numbers of odd and even Galls in the box so pristability of 3 odd balss - Probability of 3 even bally and number of cares = p (say) probability of zeven balls = probability of. 2odd & one even ball number of cases =q (say one odd ... Probability of having odd sum from 3 bully pta p+b+q1q 2 in I

(V) We have to choose three balls from the eet of balls ho 7,8 numbers B-{5, 6, having 8,9} Choosing 3 balls from B with no 5 in lt. in ways number of coays we choose balls from B to that 5 is there is 5 c₂-4e3 = 6 way. 4e3 So uez .: Probability of choosing balls.coith smallst nimber 6 5 in il 6 56 a 3 28 863