## Archive for May, 2015

## Problem 745

An experiment consists of dealing 8 cards from a standard 52- card deck. What is the probability of being dealt 8 hearts?

**Solution:-**

P(E) =

Let the sample space S be the set of all 8-card hands that could be selected from a 52-card deck. Since the order in a hand does not matter, the number of possible selections is C_{52,8}.

The number of element in S is

n(s) = C_{52,8 }= 752538150

Let the event E be the set of all 8-card hands that could be selected from the 13 hearts in the deck. Again , the order does not matter so the number of possible selection is c_{13,8}. The number of element in E is n(E) = c_{13,8} = 1287

Thus , the probability of being dealt 8 hearts is

0.000002

## Problem 744

In a family with 9 children, excluding multiple births, what is the probability of having 8 boys and 1 girl, in any order? As assume that a boy is as likely as a girl at each birth.

**Solution:-**

Probability of an Arbitrary Event under an Equally likely Assumption

P(E) =

Let the sample space S be of all possible permutations of girls and boys for 9 children.

There are 2 possibilities for each of 9 children. Thus using the multiplication principle, the total number of permutation will be 2^{9}.

Therefore, the number of element is S is

n(s) =2^{9}

= 512

Let the event E be the set of all element that correspond to the outcome “having 8 boys and 1 girl”. Since the could be 1^{st}, 2^{nd} , or 3^{rd}, there are 9 ways that 8 boys and 1 girl occur with 9 children. Therefore the number of element in E is

n(E) = 9

Thus , the probability of having 8 boys and 1 girl is

P(E) =

=

## Problem 743

In a family with 8 children, excluding multiple births, what is the probability of having 8 girls?

Assume that a girl is as likely as a boy at each birth.

**Solution:-**

Let the sample space S be of all possible permutations of girls and boys for 8 children.

There are 2 possibilities for each of 8 children. Thus using the multiplication principle, the total number of permutation will be 2^{8}. Therefore, the number of element is S is

n(S) = 2^{8} = 256

Let the event E be the set of all element that correspond to the outcome “having 8 girls”.

Since there is only one way that 8 girls out of 8 children can occur, there is only 1 element in the E.

n(E) = 1

Thus, the probability of having 8 girls is

P(E) =

=

## Problem 742

An experiment consists of drawing 1 card from a standard 52-card deck. What is the probability of drawing an 8?

**Solution:-**

A standard deck of 52 cards has four 13-card suits: diamonds, hearts, clubs and spades. The diamonds and hearts are red, and the clubs and spades are black. Ecah 13-card suit contains cards numbered from 2 to 10, a jack, a queen, a king and an ace. The jack, queen, and king are called face cards.

Drawing 1 card implies that each simple event in sample space S is as likely to occur as any other. Therefore, the probability of an arbitrary event E in S is given by the formula below.

P(E)=

The sample space S is the set of the 52 cards in a standard deck.

n(S) = 52

Let Q = the card is an 8. There are 4 element Q in the sample space.

n(Q) = 4

Substitute the value for n(Q) and n(S).

P(Q) =

=

Therefore , the probability of drawing an 8 is .

## Problem 741

Find the sum of the first 20 terms of the sequence: -1, 1, -1, 1, -1, …

**Solution:-**

Find a_{n} by using the explicit formula for geometric sequences: a_{n }= a_{1}r^{n-1} . Then use the formula for the sum of a finite geometric series:

S_{n} =

0.

## Problem 740

Write the geometric sequence that has four geometric means between 1and 1,024.

**Solution:-**

a_{n} and a_{1 } are given. Find the common ratio by using the explicit formula for geometric sequences: a_{n }= a_{1}r^{n-1} .

1, 4, 16, 64, 256, 1,024

## Problem 739

Find the next term of the sequence: -600, 300, -150, …

**Solution:-**

Find the common ratio and multiply the previous term by that value.

75

## Problem 738

Find the sum of the first 25 terms of the arithmetic series: -5, -4, -3, -2, -1, …

**Solution:-**

Find a_{n} by using the explicit formula for arithmetic sequences:

a_{n} = a_{1} + (n-1)d.

a_{1 }= -5

a_{25} = 19

The use the formula for the sum of an arithmetic series:

S_{n} = (a_{1} + a_{n})

175

## Problem 737

Find the sum of the first five terms of the arithmetic series:

a_{n} = 2n +2.

**Solution:-**

Find a_{1} and a_{5} and use the formula for the sum of an arithmetic series: S_{n} = (a_{1} + a_{n})

a_{1 }= 2* 1 + 2 = 4

a_{5} =2*5 +2 = 12

S_{n }= 40

## Problem 736

Find the 100^{th} term of the sequence: -4, -2, 0, 2, ….

**Solution:-**

Find the common difference and use the explicit formula for arithmetic sequences: .

a_{n} = a_{1} + (n-1)d.

a_{1 }= -2

n = 100

d = 2

194