mathefaro:
Actually, I think you guys have been looking at the question from a wrong angle. Something tells me that he meant (1/2)x or x/2 and (1/3)x or x/3 instead of the 1/2x and 1/3x respectively in the question.
My thoughts though

lol.
Modified*
You are very correct. Abeg solve it.


Two dice are thrown. What is the probability that the sum of the numbers is divisible by 3?

Cc thankyoujesus, shyneptune.

Geofavor:

lol. The question is correct. They got the answer correct also.


Two dice are thrown. What is the probability that the sum of the numbers is divisible by 3?

Cc thankyoujesus, shyneptune.

possible outcomes (1,2) (1,5) (2,1) (2,4) (3,3) (4,2) (5,1).
Probability=7/36

thankyouJesus:
possible outcomes (1,2) (1,5) (2,1) (2,4) (3,3) (4,2) (5,1).
Probability=7/36

(6,3), (3,6), (5,4), (4,5) (6,6)

12/36 = 1/3.... I just got it. Thanks for the eye opener.

Btw, is there a quicker method?

aj4emmanuel:
(2+x)/4x > (2+3x)/6x
6x(2+x) > 4x(2+3x)
12x + 6x^2 > 8x + 12x^2
0 > 12x^2 - 6x^2 + 8x - 12x
0 > 6x^2 - 4x
6x^2 - 4x <0
(2x)(3x-2) <0
2x <0 or 3x - 2 <0
x <0 ,3x <2
x <0 , x <2/3
2/3 >x< 0

thanks. There was a mistake in the question. Here is the question. (1/2)x + 1/4 > (1/3)x + 1/2

Teams P and Q are involved in a game of footbal. What is the probability that the game ends in a draw?

Geofavor:

(6,3), (3,6), (5,4), (4,5) (6,6)

12/36 = 1/3.... I just got it. Thanks for the eye opener.

Btw, is there a quicker method?

oops, my bad

Geofavor:
Teams P and Q are involved in a game of footbal. What is the probability that the game ends in a draw?

Hint; win win draw.

thankyouJesus:
Hint; win win draw.

I don't understand o. I thought in a football game, we can have a total of three outcomes: win, draw, lose.

:. The probability of a draw should be 1/3. But this isn't the answer

Geofavor:

I don't understand o. I thought in a football game, we can have a total of three outcomes: win, draw, lose.

:. The probability of a draw should be 1/3. But this isn't the answer

The probability that the game ends in a draw = P(P draws the game) AND P(Q draws the game)
My thoughts

mathefaro:
The probability that the game ends in a draw = P(P draws the game) AND P(Q draws the game)
My thoughts

this would give us 1/3 x 1/3 = 1/9.

The answer is. . . not 1/9

Three boys play a game of luck in which their respective chances of winning are 1/2, 1/3, and 1/4.
What is the probability that one and only one of the boys wins the game?


Thankyoujesus, mathefaro, shyneptune

Geofavor:
Three boys play a game of luck in which their respective chances of winning are 1/2, 1/3, and 1/4.
What is the probability that one and only one of the boys wins the game?


Thankyoujesus, mathefaro, shyneptune

ab'c' + a'bc' + a'b'c.
Where a',b' and c' are losing.
Losing=1-winning.

Geofavor:
Teams P and Q are involved in a game of footbal. What is the probability that the game ends in a draw?

cc agentofchange

thankyouJesus:
ab'c' + a'bc' + a'b'c.
Where a',b' and c' are losing.
Losing=1-winning.

I have done this but the past question said I failed it. This will give us 11/24.

They said the answer is 1/24

options

a) 1/24 .b) 1/12. C) 11/24. D) 23/24

.

.

aj4emmanuel:
x/2 + 1/4 > x/3 + 1/2
(4x+2)/8 > (2x+3)/6
6(4x+2) > 8(2x+3)
24x + 12 > 16x + 24
24x - 16x > 24 - 12
8x > 12
x > 12/8
x > 3/2

the answer was given to be x < 3/2... If you have a math past question. Pls check year 2010, No. 18.

I'm beginning to think my past question has some of its answers to be wrong.

aj4emmanuel:
P win, Q win, P lose, Q lose, then draw,
1/5

naah.

.

.

aj4emmanuel:
i see no reason why d sign should change to <. d sign will change only if u are dividing a number with negative sign

exactly

aj4emmanuel:
please wat are d option?

the answer was given to be 1/2

.

In how many ways can a delegation of 3 be chosen from 5 men and 3 women, if at least 1 man and 1 woman must be included?


Thankyoujesus, mathefaro, shyneptune, umartins1 etc.

Shouldn't this be 5C₂ x 3C₁ + 5C₁ x 3C₂ ?

Geofavor:


In how many ways can a delegation of 3 be chosen from 5 men and 3 women, if at least 1 man and 1 woman must be included?


Shouldn't this be 5C₂ x 3C₁ + 5C₁ x 3C₂ ?

I'm beginning to forget maths ooooo

Lolz. I think so that's it..... Is your books saying
otherwise?


Modified. ... I just took a second look at the question again. I don't think that's the right method. .. cos a man and a woman must be included. ....

Solving.......

ShyNeptune:

I'm beginning to forget maths ooooo

Lolz. I think so that's it..... Is your books saying otherwise?

in fact, the thing tire me. This my past questions say this is what it should be: ( 5C₃ + 3C₁ ) + ( 3C₂ + 5C₁ )


btw, check this one out: n + 1C₃ = 4( nC₃ ) .... Find n

@ Geofavor
I'm getting what you're getting
45

ShyNeptune:
@ Geofavor
I'm getting what you're getting
45

i'm going to burn this past questions book if this happens again. Check out the second question.

Geofavor:


btw, check this one out: n + 1C₃ = 4( nC₃ ) .... Find n

1C3 does not exist.... at least for O'Level

Geofavor:

i'm going to burn this past questions book if this happens again.

Lwkmd.

ShyNeptune:

1C3 does not exist.... at least for O'Level

the 1 and n are together. As in (n + 1)C₃

In how many ways can 6 coloured chalks be arranged if 2 are of the same colour?
A) 60. B) 120. C)240. D)360