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.

tertre Sep 15, 2016

#1**+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.

geno3141 Sep 16, 2016

#1**+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