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Strike441 [17]
3 years ago
15

Donald has nine one-day passes to Dizzyworld. He can go alone and use one, or he can take a friend and use two. If he visits eve

ry day until he uses all the passes, in how many different ways can he use them? Using a two day for four days and then going alone at the end (2,2,2,2,1) is different from reversing the order (1,2,2,2,2).
I don't know the answer to this question.
Mathematics
2 answers:
Nata [24]3 years ago
8 0
There is 5 umbers in the sequence so the 1 can go in 5 different places
Mazyrski [523]3 years ago
7 0
(1,2,2,2,2),(2,1,2,2,2)
Observe, the 1 can go in 5 diff places
so 5 ways
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An urn contains three white balls and two red balls. The balls are drawn from the urn, oneat a time without replacement, until a
Tpy6a [65]

Answer:

p ( X = 1 ) = 0.6 , p ( X = 2 ) = 0.3 , p ( X = 3 ) = 0.1

Verified

E ( X ) = 1.5

Step-by-step 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

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                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:

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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

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