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) = \frac{n(E)}{n(S)}

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 C52,8.

The number of element in S is

n(s) = C52,8  = 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 c13,8. The number of element in E is n(E) = c13,8 = 1287

Thus , the probability of being dealt 8 hearts is

\approx  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) =\frac{n(E)}{n(S)}

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 29.

Therefore, the number of element is S is

n(s)  =29

= 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 1st, 2nd , or 3rd, 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) = \frac{n(E)}{n(S)}

= \frac{9}{512}

 

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 28. Therefore, the number of element is S is

n(S) = 28 = 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) = \frac{n(E)}{n(S)}

= \frac{1}{25}

 

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)=\frac{n(E)}{n(S)}

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) = \frac{n(Q)}{n(S)}

=\frac{4}{52}

Therefore , the probability of drawing an 8 is \frac{1}{13}.

 

Problem 741

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

 

Solution:-

 

Find  an  by using the explicit formula for geometric sequences: an = a1rn-1   . Then use the formula for the sum of a finite geometric series:

Sn =  \frac{a_{1}-a_{1}r^{n}}{1-r}

0.

 

 

Problem 740

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

 

Solution:-

 

an and  a1  are given. Find the common ratio by using the explicit formula for geometric sequences:  an = a1rn-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 an by using the explicit formula for arithmetic sequences:

an = a1 + (n-1)d.

a1 = -5

a25 = 19

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

Sn = \frac{n}{2}(a1 + an)

175

 

Problem 737

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

an = 2n +2.

 

Solution:-

 

Find a1 and  a5  and use the formula for the sum of an arithmetic series:    Sn = \frac{n}{2} (a1 + an)

a1 = 2* 1 + 2 = 4

a5 =2*5 +2 = 12

Sn = 40

 

Problem 736

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

 

Solution:-

 

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

an = a1 + (n-1)d.

a1  = -2

n = 100

d = 2

194