Write the probability distribution of the defective bulbs drawn in three draw when it is known that in a lot of 10 on an average 3 are defective
Answers
your answer will be 3.333333 because.
total no of bulbs be 10
on an average defective bulbs are 3
probability formula:sum of observation
____________
total no of observations
3/10
=3.333333
Hola mate
Here is your answer -
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