A 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?

Question

anonymous · Answer

@imqwerty

anonymous · Answer

Hey there.
First, find p and q. Like I showed in your previous posted question.
p is the probability of finding a defective and q is of not finding a defective one.
p=5/15=1/3
q=1-p=1-1/3=2/3
np=15*1/3=5
nq=15*2/3=10
Both np and np are less than 5, hence a binomial expansion is to be used.$$P(X \le2)=P(X=0)+P(X=1)+P(X=2)$$
Basing from the original formula$$P(X=x)=(nCx)*p^x*q ^{n-x}$$
Try it out and tell me what you get.

anonymous · Answer

Any difficulties @Devanshi?

anonymous · Answer

Why did we multiply np and nq ?

anonymous · Answer

To know weither we use Binomial or Normal approximations to solve the problem.

welshfella · Answer

I knew it might be a binomial distribution bu I couldn't remember the details..

anonymous · Answer

If either or both np and nq are less than 5, then  a Binomial Distribution is to be used.

welshfella · Answer

@Hoslos - don't you just check on the value of np?  
correct me if i'm wrong

welshfella · Answer

oh either / or  OK

anonymous · Answer

@Devanshi ?