Question

If 10% of the pens manufactured by a company are defective, find the

probability that a box of 12 pens contains: (i) exactly 2 defective pens. (ii) at

least 2 defective pens. (iii) No defective pen.​

Answer 1

Given:

• Total number of defective pens,(p) = 10%
• Number of trials = 12

To Find:

• Probability of exactly two defective pens.
• Probability of at least two defective pens
• Probability of no defective pens.

Solution:

• Probability of no defective pens, (q) = 1 - p
• For exactly two defective pens:  
• p(x = 2)  = $12(_C)_{2}*(\frac{1}{10})^2 *(\frac{9}{10})^{10}$
• p(x = 2)  = $\frac{12*11*10!}{2*10!}*\frac{1}{100}*(\frac{9}{10})^{10}$ = $\frac{66}{100} *(\frac{9}{10})^{10}$
• p(x = 2) = 0.66*0.34
• p(x =2) = 0.2244
• For at least two defective pens:
• p(x ≥ 2) = 1 - [p(x=0)+p(x=1)]
• p(x ≥ 2) = 1 - [$12(_c)_0*(\frac{1}{10})^0 *(\frac{9}{10})^{12} +12(_c)_1*(\frac{1}{10})^{10} *(\frac{9}{10})^{11}$ ]
• p(x ≥ 2) = 1 - [$(\frac{9}{10})^{12}+12*\frac{9^{11}}{10^{12}}$ ] = 1-[0.376
• +0.282]
• p(x ≥ 2) = 1-0.0658
• p(x ≥ 2) = 0.9342
• For no defective pens:
• p(x = 0) = $12(_c)_0*(\frac{1}{10})^0 *(\frac{9}{10})^{12}$ = 0.282

Probability of exactly two defective pens = 0.2244

Probability of at least two defective pens = 0.9342

Probability of no defective pens = 0.282

Answer 2

Probability that Exactly two will be defective  is about 0.23

Probability that  at least two will be defective  is about 0.341

Probability that  No defective pen   is about 0.282

Given:

• Probability of defective pen = 10%
• 12 Such Pens manufactured

To Find:

• Probability that
• (i) exactly two will be defective
• (ii) at least two will be defective
• (iii) No defective pen.​

Solution:

P(x) = ⁿCₓpˣ(1-p)ⁿ⁻ˣ

n = 12

p =10% = 10/100 = 1/10

1 - p = 1- 1/10  = 9/10

Step 1:

Probability that Exactly two will be defective  P(2)

P(2) = ¹²C₂(1/10)²(9/10)¹²⁻²

P(2) = 66 * 9¹⁰ / 10¹²

P(2) ≈ 0.23

Step 2:

Probability that  at least two will be defective  1 - P(0) - P(1)

P(0) = ¹²C₀(1/10)⁰(9/10)¹²⁻⁰

P(1) =  ¹²C₁(1/10)¹(9/10)¹²⁻¹

1 - P(0) - P(1) ≈ 0.341

Step 2:

Probability that  No defective pen =  P(0)

P(0) = ¹²C₀(1/10)⁰(9/10)¹²⁻⁰

P(0) ≈  0.282

Probability that Exactly two will be defective  is about 0.23

Probability that  at least two will be defective  is about 0.341

Probability that  No defective pen   is about 0.282