Question

Box of 15 gadgets is known to contain 5 defective gadgets if 4 gadgets are drawn at random what is the probability of finding not more than 2 defective gadgets

Answer 1

To solve this problem, we make use of the Binomial Probability equation which is mathematically expressed as:

P = [n! / r! (n – r)!] p^r * q^(n – r)

where,

n = the total number of gadgets = 4

r = number of samples = 1 and 2 (since not more than 2)

p = probability of success of getting a defective gadget

q = probability of failure = 1 – p

 

Calculating for p:

p = 5 / 15 = 0.33

So,

q = 1 – 0.33 = 0.67

 

Calculating for P when r = 1:

P (r = 1) = [4! / 1! 3!] 0.33^1 * 0.67^3

P (r = 1) = 0.3970

 

 

Calculating for P when r = 2:

P (r = 2) = [4! / 2! 2!] 0.33^2 * 0.67^2

P (r = 2) = 0.2933

 

Therefore the total probability of not getting more than 2 defective gadgets is:

P = 0.3970 + 0.2933

P = 0.6903

 

Hence there is a 0.6903 chance or 69.03% probability of not getting more than 2 defective gadgets.