RI S2B Binomial Distribution tutorial (Solns)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 _______________________________ Tutorial S2B: Binomial Distribution Page 1 of 11 Tutorial S2B: Binomial Distribution Section A (Discussion Questions) 1 State, with a reason, whether a binomial distribution could be used in each of the following problems. If binomial distribution is an acceptable model, define the random variable clearly and state its parameters. (a) A gambler has a biased coin for which the probability of obtaining a head in any toss is 0.56. Find the probability of getting 6 heads if he tosses the coin 8 times. (b) A fair coin is spun until a head occurs. Find the probability that 8 spins are necessary, including the one on which the head occurs. (c) A jar contains 49 balls numbered 1 to 49. Six of the balls are drawn at random without replacement. Find the probability that four out of the six balls drawn have an even score. Solution: (a) Since there are 8 independent trials (tosses of coins), each toss either yields a head or tail and the probability of getting a head stays constant at 0.56 from toss to toss, binomial distribution is an acceptable model. Let X be the no. of heads obtained, out of 8 tosses.       6 2~ B 8, 0.568P 6 0.56 0.44 0.1676XX      (b) No. For binomial distribution to be used, it should consist of n independent spins but in this case, there are no fixed n number of spins. (c) No. For binomial distribution to be used, the draws should be independent of each other, which is not the case here since the balls are drawn with no replacement. Note: Probability of drawing an even-numbered ball is constant regardless of the draw. Eg P(drawing an even-numbered ball in 1st draw) = 2449 P(drawing an even-numbered ball in 2nd draw) = P(drawing an even-numbered ball in 2nd draw and odd-numbered ball in 1st draw) + P(drawing an even-numbered ball in 2nd draw and even-numbered ball in 1st draw) 25 24 24 23 2449 48 49 48 49     Similarly, P(drawing an even-numbered ball in 3rd draw) =… = P(drawing an even-numbered ball in 6th draw) = 2449

Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________________ ________________________________ Tutorial S2B: Binomial Distribution Page 2 of 11 2 Given ~ B 10, 0.2X , find (i) ; (ii) ; (iii) ; (iv) P(4 8)X  ; (v) P(2 6)X  ; (vi) E X ; (vii) Var X . [ (i) 0.107 (ii) 0.678 (iii) 0.0328 (iv) 0.121 (v) 0.618 (vi) 2 (vii) 1.6] (i) and (ii) involves direct application of graphing calculator. (iii) = 1 P 4X  = 0.0328 (iv) P 4 8 P 8 P 3X X X      0.121 (v) P 2 6 P 5 P 1X X X      0.618 (vi) E X np 10 0.2 2   (vii) Var 1X np p  10 0.2 1 0.2 1.6    3 A die is biased such that the probability of getting a six is 0.18. If the die is thrown ten times, find (i) the probability that the number of sixes that turn up is odd, (ii) the expectation and variance of the number of sixes that turn up. [(i) 0.494, (ii) 1.8, 1.476] (i) Let X be the number of sixes obtained, out of 10 throws. ~ B 10, 0.18X   P is odd P 1 +P 3 P 5 +P 7 +P 9 = 0.494 3 s.f. X X X X X X       (ii)   E = 10 0.18 = 1.8 X np    Var 1 = 10 0.18 1 0.18 = 1.476 X np p   Expectation and variance of number of sixes that turn up is 1.8 and 1.476 respectively. P( 0)X P( 2)X P( 5)X P( 5)X

Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________________ ________________________________ Tutorial S2B: Binomial Distribution Page 3 of 11 4 Two fair dice are thrown. Find the probability that the total score is two-figured. If five people were each to throw two dice, find the probability that exactly three of them will get a two-figured total score. [ 1 6, 0.0322] Total score is two-figured : 4, 6 , 5, 5 , 5, 6 , 6, 4 , 6,5 , 6, 6 . Probability that total score is two-figured = 6 1 36 6 Let Y be the number of people, out of the 5, who get a two-figured total score. 1~ B 5,6Y     P 3 0.0322 3 s.f.Y  5 RJC Prelim 1997/02/Q7(a)(modified) The probability that a patient suffering from a particular disease will be cured following a new treatment is 0.92. (i) If the treatment is given to 10 patients, show that the probability that at least 9 out of the 10 patients will be cured is 0.812, correct to 3 significant figures. (ii) This treatment is being tested at 12 hospitals where in each hospital, the treatment is given to 10 patients. Find the probability that more than 10 of the 12 hospitals will have a success rate of at least ninety percent. [(ii) 0.310] (i) Let X be the number of patients, out of 10, who will be cured following the new treatment. ~ B 10, 0.92X    P 9 P 9 +P 10 = 0.81212 = 0.812 3 s.f. Shown X X X    OR:    P 9 1 P 8 = 1 0.18788 = 0.81212 = 0.812 3 s.f. Shown X X     (ii) Let Y be the number of hospitals each with 10 patients, out of 12 hospitals, which will have a success rate of at least 90 percent. Probability of a hospital having success rate of at least 90 percent = 0.812 (from (i)) ~ B 12, 0.812Y   P 10 1 P 10 = 0.310 3 s.f. Y Y   

Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________________ ________________________________ Tutorial S2B: Binomial Distribution Page 4 of 11 6 SRJC Prelim 2007/02/Q10a According to the school rules, a student who arrives at school after 0730 hours is considered late. (i) During the school term, a boy cycles to school on 5 days each week. On any given day, the probability that he arrives after 0730 hours is 0.1. For a period of 4 weeks, calculate the probability that he is late on at least one day but not more than 3 days. When the boy has cycled to school for the nth time, the probability that he arrived at school late at least once is greater than 0.99. Calculate the least value of n. (ii) A girl travels to school by bus on 5 days a week. Over a long period of time, the variance of the number of days per week on which she is late is 0.8. Given that p denotes the probability that she arrives late and that 0.5p , evaluate p. [(i) 0.745, 44 (ii) 0.2] Let X be the number of days a student is late in 4 weeks. ~ B(20, 0.1)X P(1 3) P( 3) P( 0) = 0.745 X X X      Let Y be the number of days a student is late in n days. ~ B( , 0.1)Y n 0 P( 1) 0.99 1 P( 0) 0.99 P( 0) 0.01 (0.1) (0.9) 0.010 ln(0.9) ln(0.01) ln(0.01)43.70869ln(0.9) n Y Y Y n n n                    least value of n is 44. Let W be the number of days the girl is late in a week. ~ B(5, )W p Variance = 0.8 2 5 (1 ) 0.8 0.16 0 0.8 or 0.2 p p p p p          Since p < 0.5, p = 0.2 Alternatively using GC (table), P 0 0.01Y  n P 0Y 43 0.0108 > 0.01 44 0.00970 < 0.01 least value of n is 44.

Raffles Institution H2 Mathematics 2025 Year 6 _____________________