Probability

[kofigc]

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 to anyone who helps me

 

[wjm11]

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.

You have errors on 1i and 1ii. The other answers are correct.

1i and 1ii are not the same. 1ii has two possible scenarios: both are red or both are black. Your sample space for this problem is {RR, RB, BR, BB}. The sum of the probabilities of these four outcomes must add up to 1. P(RR) = P(BB). P(RB) = P(BR). Does that help any?

 

[soroban]

Hello, kofigc!

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:

\(\displaystyle i)\; P(\text{neither break}) \;=\; (1-0.04)\cdot(1/0.02) \;=\; (0.96)(0.98) \;=\;\frac{588}{625}\)

\(\displaystyle ii)\; P(\text{one breaks}) \;=\; (0.04)(1-0.02) \;+\; (0.02)(1-0.04) \;=\; (0.04)(0.98) \;+\;(0.02)(0.96) \;=\; 0.0392\;+\;0.0192 \;=\; \frac{73}{1250}\)

\(\displaystyle iii)\; P(\text{both break}) \;=\; (0.04)(0.02) \;=\;\frac{1}{1250}\)

My problem is that with the above answers I get 71% for the 3 questions, . How did you get 71%?

Your answers are correct!


. . \(\displaystyle \begin{array}{cccc} P(\text{neither break}) &=& 0.9408 \\ P(\text{one breaks}) &=& 0.0584 \\ P(\text{both break}) &=& 0.0008 \\ \hline && 1.0000 \end{array}\)