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An element is randomly chosen from among the first 15   rows of Pascal's Triangle. What is the probability that the value of the element chosen is 1 ?

Note: The 1 at the top is often labelled the "zeroth" row of Pascal's Triangle, by convention. So to count a total of 15 rows, use rows 0 through 14.

 Sep 15, 2016

Best Answer 

 #1
avatar+17776 
+10

Row 0 has only 1 element:  1

Row 1 has two elements:    1 1

Row 2 has three elements:  1 2 1

Row 3 has four elements:    1 3 3 1

Following this pattern, row 14 has 15 elements.

So, the total number of elements is:  1 + 2 + 3 + ... + 15.

You can find the sum of these numbers by using the formula:  Sum  =  N(F + L) / 2

     where N = 15, the number of numbers;  F = 1, the first number;  L = 15, the last number.

     Sum  =  15(1 + 15) / 2  =  120

How many of these numbers are 1?  There is only one element in row 0, and it is a 1.  For all the other rows, both the first element and the last element are 1; thus 2 1's in each of those rows.

1 + 2 x 14  =  29.

So, the probability of randomly selecting the number 1 is  29 / 120.

 Sep 16, 2016
 #1
avatar+17776 
+10
Best Answer

Row 0 has only 1 element:  1

Row 1 has two elements:    1 1

Row 2 has three elements:  1 2 1

Row 3 has four elements:    1 3 3 1

Following this pattern, row 14 has 15 elements.

So, the total number of elements is:  1 + 2 + 3 + ... + 15.

You can find the sum of these numbers by using the formula:  Sum  =  N(F + L) / 2

     where N = 15, the number of numbers;  F = 1, the first number;  L = 15, the last number.

     Sum  =  15(1 + 15) / 2  =  120

How many of these numbers are 1?  There is only one element in row 0, and it is a 1.  For all the other rows, both the first element and the last element are 1; thus 2 1's in each of those rows.

1 + 2 x 14  =  29.

So, the probability of randomly selecting the number 1 is  29 / 120.

geno3141 Sep 16, 2016

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