[kofigc]

1) Two cards are drawn in succession from a well shuffled pack of cards (without replacement).
Find the probability that:
I) both cards are red
ii)both cards are the same colour
iii) the cards are different colours

2) A box contains 7 red balls and 11 black calls. Two balls are taken out one after the other without replacement. What is the probability that both balls are black?

3) Of two air-conditioners X and Y in a computer lab, X has a probability 0.04 of breaking down on any one day and Y has a probability 0.02. On the first day os summer when both air conditioners are called into use, what is the probability that:
I) Neither will break down
ii) Exactly one will break down
iii) Both will break down

[Unknown008]

1)
i) P(both red) = P(1st red) x P(2nd red) =

the number of red cards and the total number of cards decreases after the first draw. Can you find the probability now? And post your attempts at the other questions first.

[kofigc]

Hey, thank you for the reply! I forgot to post up the attempts I've taken - oops!
Hi, I have a few questions on an excel file that I have to complete regarding probability. Here are the questions:
1) Two cards are drawn in succession from a well shuffled pack of cards (without replacement).
Find the probability that:
I) both cards are red
ii)both cards are the same colour
iii) the cards are different colours

2) A box contains 7 red balls and 11 black calls. Two balls are taken out one after the other without replacement. What is the probability that both balls are black?

3) Of two air-conditioners X and Y in a computer lab, X has a probability 0.04 of breaking down on any one day and Y has a probability 0.02. On the first day os summer when both air conditioners are called into use, what is the probability that:
I) Neither will break down
ii) Exactly one will break down
iii) Both will break down

Here are my answers:
1i) P(1st red) = 1/2, P(2nd red) = 25/51 so P(both red) = 1/2 *25/51 = 25/102

ii) 25/102 (same concept as above)

iii) P(1st red) = 1/2, P(2nd black) = 26/51 so P(different colours) = 1/2 * 26/51 = 13/51

2) p(1st ball black) = 11/18, P(2nd ball black) = 10/17 so P(both balls black) = 11/18 * 10/17 = 55/153

3i) P(neither break) = (1-0.04) * (1/0.02) = 0.96*0.98 = 588/625

ii) P(one breaks) = 0.04 * (1-0.02) + 0.02 * (1-0.04) = 0.04*0.98 +0.02*0.96 = 0.0392+0.0192 = 73/1250

iii) P(both) = 0.04 * 0.02 = 1/1250

My problem is that with the above answers I get 71% for the 3 questions, I would like to know where and what I'm doing wrong.
Thank you once again :)

[Unknown008]

1) ii)

Not quite right. :)

P(both are same colour) = P(both are red OR both are black) = P(both Red) + P(both black)

iii) P(different colours) = P(1st red, 2nd black OR 1st black, 2nd red) = P(1st red, 2nd black) + P(1st black, 2nd red)

That's it! Post what you get for those.

[kofigc]

Ah! I see what I've done wrong now and once again thank you for the answer (very fast I must add)!
1ii) 25/51?
iii) 26/51?
Again, I really appreciate the time and advice you've given, thank you!

[Unknown008]

Well done! :)

[kofigc]

Haha! :D
Can't thank you enough, those have been bugging me for days!
One last time: thank you :D