Homework Help: math

Posted by michele on Friday, February 25, 2011 at 5:54pm.

two teams meet in playoff series at the end of the regular season. Team A won 55 of 81 games played in its home stadium during the regular season, while Team B won 48 of 81 games played in its home staduim. The first two games of the series will be played in Team A's home stadium the next two games in Tean B's home stadium. In the adsence of any other information, which expression is equal to the probability that Team A will win the first four games in a row?

math - michele, Friday, February 25, 2011 at 6:06pm
a solid cube of a aluminum has sides 9cm long. A metallurgist would like to melt the cube down and use all the molten aluminum to make three smaller, solid congruent cubes. What should the lentgh, in centimeters, of a side of one of the smaller cubes?
math - michele, Friday, February 25, 2011 at 6:33pm
a box contains 5 red and 4 blue balls. In how manys ways can 4 balls be chosen such that there are at most 3 balls of each colour.
math - MathMate, Friday, February 25, 2011 at 7:04pm
Michelle, please do not piggy-back questions on your own posts. It may slow down responses to your questions because not all tutors are ready to attack all the problems in one breath.

1.
Assuming the statistics are independent of the visiting team, then probability of team A winning in home turf is P(A)=55/81, and winning as a visitor of team B is P(B)=(81-48)/81=33/81.
So winning all four games,
P = P(A)*P(A)*P(B)*P(B)

2.
Total volume = 9³ cm³ = 729 cm³.
Side of each of 3 new cubes
= cube root of (729 cm³ /3) cm
= cube root of (243) cm.

3.
Total number of ways to pick 4 balls out of 9
= C(9,4) = 9!/(4!5!)=126
Total number of ways to pick 4 blue balls = C(4,4) = 1
total number of ways to pick 4 red balls
= C(5,4) = 5

Number of ways of not choosing 4 red nor 4 blue = (126 - 1 - 5) = 120

Probability of not choosing 4 balls of the same colour is therefore ...?
math - michele, Friday, February 25, 2011 at 7:53pm
Cuts of beef/High fat/Low fat/ total
Flank steaks/ 74 / 386 / 460
Rump roasts/ 258 / 142 / 400
Total /332 /528 / 860

A USDA inspector is grading cuts of beef at a meat packing plant. If a piece of beef is selected at random, what is the probability that it will be a flank with high fat content?
math - MathMate, Friday, February 25, 2011 at 8:39pm
Flank with high fat = 74 pieces
Total number of pieces = 860
So P(Flank HF) = 74/?

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