SAJC Check your Understanding Binomial dist solutions

Binomial Distribution: Check your Understanding Section 1: Assumptions 1. ACJC Prelim 8865/2018/Q7part A candy factory manufactured a large amount of pastilles daily and their candies are randomly packed in boxes of 20. The probability of selecting an orange-flavoured pastille to be packed into a box is 0.25. The random variable X is the number of orange-flavoured pastilles in a box of 20 pastilles. State an assumptions that is needed for X to be modelled by a binomial distribution. [2] Soln: A pastille being orange-flavour is independent of any other pastille being orange flavor. Or The flavour of each pastille must be independent of the flavour of any other pastille. 2. AJC Prelim 8865/2018/Q7part On average 7% of a certain brand of kitchen lights are faulty. The lights are sold in boxes of 12. State, in context, two assumptions needed for the number of faulty lights in a box to be well modelled by a binomial distribution. [2] Soln: A kitchen light being faulty is independent of any other kitchen light being faulty. The probability of a light being faulty is constant at 0.07 for every lights in a box. 3. CJC Prelim 8865/2018/Q6part As part of Singapore’s aim to be a Smart Nation in 10 years, hawker centres are encouraged to go cashless. An initial trial of cashless payment in hawker centres shows that 1 in 5 customers uses cashless payment. A random sample of 12 customers is selected and the number of customers who use cashless payment is denoted by the random variable C. State, in context, an assumption needed for C to be well modelled by a binomial distribution. [1] Soln: A customer who uses cashless payment is independent of any other customers who use cashless payment. The probability of customers using cashless payment is constant at 0.2 for every customers. Section 2: Find Probability 1. YJC Promo 8865/2018/Q6 A health agency launches a ‘2+2’ campaign to encourage people to eat two servings of fruits and two servings of vegetables per day. In a particular school, 20% of students eat at least 2+2 fruits and vegetables per day. 18 students in the school are selected at random and the number of students who eat at least 2+2 fruits and vegetables per day is denoted by X. (i) Find the probability that more than 3 of them eat at least ‘2+2’ fruits and vegetables per day, [2]

(ii) Find the probability that between 2 and 7 of them eat at least ‘2+2’ fruits and vegetables per day, [2] Answer: (i) 0.499, (ii) 0.677 Soln: (i)  ~ B 18,0.2X ( 3) 1 ( 3) = 0.499 P X P X    (ii)      P 2 7 P 6 P 2 0.9481290011 0.271348775 0.677 (3 s.f.) X X X        2. RI Prelim 8865/2018/Q8part A dental clinic sees 50 patients each day for a total of 6 days each week, and is closed on Sunday. On average, 30 % of the patients are eligible for a dental subsidy after treatment, and the eligibility of a patient for dental subsidy is independent of another patient. (i) Find the probability that, on a randomly chosen day, at least 20 patients are eligible for the dental subsidy. [2] (ii) Find the probability that, in a randomly chosen week, at least 85 and at most 95 patients are eligible for the dental subsidy. [3] Answer: (i) 0.0848, (ii) 0.512 Soln: (i) Let X be the random variable denoting the number of patients that are eligible for the dental subsidy out of 50 patients, X ~ B (50, 0.3) P (X ≥ 20) = 1  P (X < 20) =1  P (X  19) = 0.0848 (3 sf ) (ii) Let Y be the random variable denoting the number of patients that are eligible for the dental subsidy out of 300 patients Y ~ B (300, 0.3) P (85  Y  95) = P (Y  95)  P (Y  84) = 0.512 (3 sf ) 3. MJC Promo 8865/2018/Q4 In a factory that produces a large number of sweets per day, a proportion, 0.15, of the sweets produced is red. A random sample of 10 sweets is taken on a particular day. The random variable X denotes the number of red sweets in the sample. (i) Find the probability that there are at least 2 red sweets in the sample. [2] (ii) Find the probability that the sample has at most 5 red sweets given that it has at least 2 red sweets. [3] Answer: (i) 0.456, (ii) 0.997 Soln: (i) Let X be the random variable denoting the number of red sweets out of 10 sweets.

 ~ 10,0.15XB P( 2) 1 P( 2) 1 P( 1) 0.45570 0.456 (3 s.f ) X X X          (ii)       P 5 2 P 5 2 P( 2) P(2 5) P( 2) P 5 P( 1) 0.45570 0.99862 0.54430 0.45570 0.99695 0.997 (3 s.f ) XX XX X X X XX              4. SAJC Prelim 8865/2018/Q8 Tom owns a cheese tart specialty shop. On a daily basis, he prepares enough ingredients to bake exactly 500 cheese tarts a day. On average, Tom sells 80% of his cheese tarts per day. Find the probability that Tom sells between 300 and 420 cheese tarts inclusive given that he sells at least 380 cheese tarts in any randomly chosen day. [3] Answer: 0.990 Soln: Let A be the random variable denoting the number of cheese tarts that Tom sells out of 500 tarts. ~ (500,0.8)AB Required Probability P(300 420 | 380) P(300 420 380) P( 380) P(380 420) P( 380) AA AA A A A            P( 420) P( 379) 1 P( 379) 0.99049 0.012256 1 0.012256 0.990 AA A       

Section 3: Find unknow p 1. JJC Prelim 8865/2018/Q8part A market research was conducted in a town where a large number of households were asked if they subscribe to fibre broadband internet services. The probability that a household subscribes to fibre broadband internet services is found to be p. A random sample of 30 households from a particular block of flats in the town was surveyed. Given that the probability that no household subscribes to fibre broadband internet services in the sample is 0.05, determine the value of p. [2] Answer: 0.0950 Soln: Let X be the random variable denoting the number of households surveyed subscribe to fibre broadband internet services out of 30. Then B(30, )Xp Given  P 0 0.05X  30 0 30 0 30 1 30 1 30 (1 ) 0.05 (1 ) 0.05 1 (0.05) 1 (0.05) C p p p p p     0.095034 0.0950 (3 s.f.)p   2. MI Prelim 8865/2018/Q7 A company supplies a particular type of mechanical seals, called gaskets, to an automotive manufacturer. They are supplied in batches of 100. On average, the proportion of rejected gaskets is p. The probability that there are at least two gaskets rejected in a randomly chosen batch is 0.02. Write down an equation involving p and hence find the value of p. [3] Answer: 0.00216 Soln: Let X be the random variable denoting the number of gaskets that are rejected, out of 100.  ~ B 100,Xp                 100 9901 100 99 P 2 1 P 1 0.02 1 P 0 P 1 0.02 100 1001 1 1 0.02 001 0.98 1 100 1 0 XX XX p p p p p p p                               Using GC, 0.00216 (3 sf)p

3. VJC Prelim 8865/2018/Q8 On average 100 %p of a certain company’s pea seeds germinate. The pea seeds are sold in trays of 24. The probability that 15 or 16 pea seeds germinate in a tray is 0.086550 correct to 6 decimal places. Find the value of p to a suitable degree of accuracy, given that 0.5p . [3] Answer: 0.7942 Soln: Let X be the random variable denoting the number of seeds that germinate out of a tray of 24.  ~ B 24,Xp               15 9 16 824 24 15 16 P 15 P 16 0.086550 1 1 0.086550 XX C p p C p p         By GC, 0.794233 0.7942 or 0.476512 (r ej 0.5)p p p    4. HCI Prelim 2008 The probability of obtaining a ‘head’ from a biased coin is p. The coin is tossed 8 times and the number of ‘tails’ obtained is denoted by T. Find p, given that     2 E( )Var T T . [4] Answer: 8 9p Soln: T ~ B(8, 1 – p) Var(T) =     22 ( ) 8(1 ) 8(1 )   E T p p p 8(1 )  pp 8 9p

Section 4: Find unknown