Answer:
p ( X = 1 ) = 0.6 , p ( X = 2 ) = 0.3 , p ( X = 3 ) = 0.1
Verified
E ( X ) = 1.5
Explanation:
Solution:-
- An urn contains the following colored balls:
Color Number of balls
White 3
Red 2
- A ball is drawn from urn without replacement until a white ball is drawn for the first time.
- We will construct cases to determine the distribution of the random-variable X: The number of trials it takes to get the first white ball.
- We have three following case:
1) White ball is drawn on the first attempt ( X = 1 ). The probability of drawing a white ball in the first trial would be:
p ( X = 1 ) = ( Number of white balls ) / ( Total number of ball )
p ( X = 1 ) = ( 3 ) / ( 5 )
2) A red ball is drawn on the first draw and a white ball is drawn on the second trial ( X = 2 ). The probability of drawing a red ball first would be:
p ( Red on first trial ) = ( Number of red balls ) / ( Total number of balls )
p ( Red on first trial ) = ( 2 ) / ( 5 )
- Then draw a white ball from a total of 4 balls left in the urn ( remember without replacement ).
p ( White on second trial ) = ( Number of white balls ) / ( number of balls left )
p ( White on second trial ) = ( 3 ) / ( 4 )
- Then to draw red on first trial and white ball on second trial we can express:
p ( X = 2 ) = p ( Red on first trial ) * p ( White on second trial )
p ( X = 2 ) = ( 2 / 5 ) * ( 3 / 4 )
p ( X = 2 ) = ( 3 / 10 )
3) A red ball is drawn on the first draw and second draw and then a white ball is drawn on the third trial ( X = 3 ). The probability of drawing a red ball first would be ( 2 / 5 ). Then we are left with 4 balls in the urn, we again draw a red ball:
p ( Red on second trial ) = ( Number of red balls ) / ( number of balls left )
p ( Red on second trial ) = ( 1 ) / ( 4 )
- Then draw a white ball from a total of 3 balls left in the urn ( remember without replacement ).
p ( White on 3rd trial ) = ( Number of white balls ) / ( number of balls left )
p ( White on 3rd trial ) = ( 3 ) / ( 3 ) = 1
- Then to draw red on first two trials and white ball on third trial we can express:
p ( X = 3 ) = p ( Red on 1st trial )*p ( Red on 2nd trial )*p ( White on 3rd trial )
p ( X = 3 ) = ( 2 / 5 ) * ( 1 / 4 ) * 1
p ( X = 3 ) = ( 1 / 10 )
- The probability distribution of X is as follows:
X 1 2 3
p ( X ) 0.6 0.3 0.1
- To verify the above the distribution. We will sum all the probabilities for all outcomes ( X = 1 , 2 , 3 ) must be equal to 1.
∑ p ( Xi ) = 0.6 + 0.3 + 0.1
= 1 ( proven it is indeed a pmf )
- The expected value E ( X ) of the distribution i.e the expected number of trials until we draw a white ball for the first time:
E ( X ) = ∑ [ p ( Xi ) * Xi ]
E ( X ) = ( 1 ) * ( 0.6 ) + ( 2 ) * ( 0.3 ) + ( 3 ) * ( 0.1 )
E ( X ) = 0.6 + 0.6 + 0.3
E ( X ) = 1.5 trials until first white ball is drawn.