Question

Write the probability distribution of the defective bulbs drawn in three draws when it is known that in a lot of 10 on an average 3 are defective

Answer 1

Answer:

Binomial distribution B(3, 0.3).

P(X=0) = 0.343

P(X=1) = 0.441

P(X=2) = 0.189

P(X=3) = 0.027

Step-by-step explanation:

Okay, so we're taking 3 bulbs and we're interested in how many of those 3 bulbs are defective.

Let X be the number of defective bulbs amongst the 3 bulbs that are selected.

Notice X takes one of the values 0, 1, 2 or 3.

The probability that an individual bulb should be defective is p=0.3, since (on average), 3 in every 10 are defective.

This "select n items, each has probability p of being (something), count the number that are (something)" scenario corresponds to a binomial distribution, B ( n, p ).

So X has a binomial distribution B( 3, 0.3 ).

Since n = 3 is so small (X only takes 4 different values), we can easily write out the distribution in detail.

P(X=0) = (1-p)³ = 0.7³ = 0.343

P(X=1) = 3p(1-p)² = 3×0.3×0.7² = 0.441

P(X=2) = 3p²(1-p) = 3×0.3²×0.7 = 0.189

P(X=3) = p³ = 0.3³ = 0.027

Answer 2

Hola mate

Here is your answer -

Kindly see the attachment...