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A box contains 10 red markers, 4 green markers, 2 yellow markers, and 4 blue markers.?

(a) With replacement, what are the chances of getting 4 red markers in a row?

(b) With replacement, what are the chances of getting 1 red marker, then a blue marker, then a yellow marker, and finally, a green marker?

(c) Without replacement, what are the chances of getting 1 red marker, then a blue marker, then a yellow marker, and finally, a green marker?

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  • 1 month ago
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    This looks like it ought to have been posted in the Mathematics section. 

     

    (a) With replacement, what are the chances of getting 4 red markers in a row?  

     

    "With replacement" means we put back each marker before randomly pulling out the next one. Half the markers (10 out of 20) are red, so the probably of getting 4 red ones in a row is 

    (1/2) (1/2) (1/2) (1/2) = 1/16 

     

     

    (b) With replacement, what are the chances of getting 1 red marker, then a blue marker, then a yellow marker, and finally, a green marker?  

     

    This is similar to part (a), except that the probabilities of different colors vary: 

    10/20 = 1/2 for red, 

    4/20 = 1/5 for green and blue, and 

    2/20 = 1/10 for yellow. 

    Because we're drawing WITH replacement, the probability of completing this sequence, in order, is 

    (1/2) (1/5) (1/10) (1/5) = 1/500 

     

     

    (c) Without replacement, what are the chances of getting 1 red marker, then a blue marker, then a yellow marker, and finally, a green marker? 

     

    WITHOUT replacement, we're drawing from a smaller set of markers at each step: 

    (1/2) (4/19) (2/18) (4/17) 

    = (4 * 2 * 4) / (2 * 19 * 18 * 17) 

    = 8 / (19 * 9 * 17) 

    = 8/2,907

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