January 17, 2007

Sum of Arithmetic / Geometric Sequence

This is RICHARD and I'm your scribe for today.

Before we started our lesson in Geometric sequences, Mr. K discussed the solutions on the on-line quiz on probability.

(1) A card is drawn randomly from a deck of 52 playing cards. What is the probability that it will be a spade or a face card?

(3) A certain partial deck of cards contains 6 red cards and 4 black cards. Two cards are chosen at random. What is the probability that both cards are black?

R = 6

B = 4

____ = 6/45


(6) The diagram shows a partial map of a certain city park, with walking paths located on the grid lines. A tourist starts at point A and randomly selects a path to point C, walking only to the south and east. What is the probability the tourist passes through point B?

The number of ways to get to point C is RRRDD which is 5!/3!2!. The number of ways to get to point B is RRD which is 3!/2!. The number of ways to get to point B over the number of ways to get to point C is 3/10

(7) It is known that about 5% of coffee cups used in a particular restaurant will be chipped. What is the probability that out of 10 cups selected at random, exactly 2 will be chipped?


C - represents the chipped
N- represents not chipped

10!/8!2! (o.o5)² (0.95)^8
= 0.0746

(8) Three coins are tossed. What is the probability that at least two heads turn up?

These are the possibilities that you can get at least two heads

HHT = 1/8

HTH = 1/8

THH = 1/8

HHH = 1/8

The sum of these events is 1/2

(10) A jar contains 5 red and 7 blue marbles. What is the probability of pulling out 2 blue marbles in a row, without replacement?

(2) A certain kind of die has four sides, labelled 1 to 4. When the die is rolled, each side has a probability of 1/4 that it will appear on the bottom. If the die is rolled 5 times, what is the probability that the side labelled “3" appears on the bottom in exactly two of the five rolls?


= 5!/ 2!3! (1/4)² (3/4)³

(4) Over the course of his basketball career, Julius succeeded in making a free throw in 75% of his attempts. In a randomly selected game, he attempted to make a free throw 12 times. What is the probability that he succeeded in 8 of the 12 attempts? (You don't need to know anything about basketball to answer this problem.)

12C8 (0.75)^8 (0.25)^4

(5) Three different names are randomly selected from the following list of five names. Max, Kim, Codie, Lee and, Alex. Determine the probability that “Kim” is one of the three names selected.

(1· 4C2)/ 5C3

= 3/10

(9) Assuming equally likely probabilities for male and female births, what is the probability that a four-child family will have at least one boy?

P(all girls) = (1/2)^4 = 1/16

P( at least 1 boy) = 1 - 1/16 = 15/16

After we had discussed about the on-line quiz, we proceed to our topic in Geometric sequences.

A mathematician named Carl Friedrich Gauss is a brilliant student. One day his teacher got angry because the class is so noisy. She gave an activity to keep them busy. She asked them to add the numbers from 1 to 100. Gauss came out with a better solution. He instantly computed the answer 5050.

t1, t1 + d, t1 + 2d, t1 + 3d, . . . . . . t1 +97d , t1 + 98d, t1 + 99d

Find the sum of the arithmetic sequence 2, 7, 12 17, . . . . S 51

S51 = 51/2 [2(2) + (51-1) 5 ]

= 51/2 [ 4 + 250 ]

=51/2 (254)

= 6477

given the geometric sequence 16, 8, 4, 2, . . . .Find the sum of this sequence up to th 10th term.

r = 1/2

t1 = 16

that's it for today!! the next scribe will be Jefferson..

No comments:

Post a Comment