The Student Room Group

more probability questions

I am stuck on probability questions again my teacher never explains any of these things and we are stuck learning it from the book so can anyone please help me.

1. Suppose the two joker cards are left in a standard deck of cards. One of the jokers is red and the other is black. A single card is drawn is from the deck of 54 cards,but not returned, and then a second card is drawn. Determine the probability of drawing the red joker or a red ace on either draw.

I know it something like 1- (Probability of not getting an ace or a club) but how do we find that number because the card is not returned. The probability of a joker is 1/54 and the aces is 2/53 if its seconds. So how would I do this.

Now i need help with this question.

2. A paper bag contains a mixture of 3 types of candy. There are ten 10 chocolate bars, seven fruit bars, and three packages of toffee. Suppose a game is played in which a candy is randomly taken from the bag, replaced, and then a second candy is drawn from the bag, if it was the same type as the one that was drawn the first time, calculate the probability of each of the following.
a)you will be able to keep the chocolate bar
b)You wont be able to keep any of them.

This one all i know for a is that the P(chocolate)=10/20
For "b" i have no clue how to do it.

Finally the last question i need help with please.

3. A basketball player has a success rate of 80% for shooting free throws.
Calculate the following probabilities.
i) She will make 3 out of 5 attempts
ii)She will miss all 5
iii)She will make the first 3 and miss the next 2.

Sorry for posting all these questions but i really need help with them because it really hard to learn from a book and your help is greatly appreciated. Unless someone can explain a very good way on how to tackle these problems or where i can get help, i will probably need help again.
Well i read the first one in your prevs post. It should be 1- (51/54)².
If not returned then 1- (51/54)*(50/53).

I'm at work so i cant read/ans the rest right now.
Reply 2
The first question i posted here is a little bit different than the one i posted in my previous thread.
Reply 3
I think this is correct.

1. P(RA OR RJ) means either P(RA, N) or P(N, RA) or P(RJ, N) or P(N, RJ)
N = neither
which is, 2/54×51/53 + 51/54×2/53 + 1/54×51/53 + 51/54×1/53
=17/159
Dont know if this is one of those questions which involves combinations ! as well :confused:
Reply 4
2.
Ok so you mean if the 2nd bar is the same as the first, then you get to keep it.
(a) P(choc, choc) = (10/20)² = 1/4
(b) 1-P(Choc, choc)-P(fruit, fruit)-P(toffee, toffee)
=121/200
Reply 5
3. p = 0.8 q = 0.2 n = 5
(i) using NcR × p^r × q^(n-r)
P(X=3) : 5C3 × 0.8³ × 0.2² = 128/625
(ii) P(X=0) :5C0 × 0.8^0 × 0.2^5 = 1/3125
(iii) 0.8³×0.2² = 64/3125

I hope these are all right but knowing my luck JonnyW (excellent statitician) will correct all my mistakes :frown:
Reply 6
Your answer for number 1 is different from my book, your answer for number 2 and 3 are right but for number 3 we havent learned that way of solving questions like that yet. We are still on conditional probability
Reply 7
What is the anwer in the book then?
Reply 8
Question 1, you can do quite similar to the question you asked before in last post.
But you have to consider 2 cases:

- 1st card is red joker or 2 card is a red ace
P1 = 1 - P(1st isn't red joker and 2nd isnt red ace)
= 1 - (53/54)(51/53)

- 1st card is red ace, or 2nd is a red joker
P2 = 1 - P2'
= 1 - (52/54)(52/53)

Then P = P1 + P2.
Well, I hope it will give you right answer :wink:
Reply 9
BCHL85
Question 1, you can do quite similar to the question you asked before in last post.
But you have to consider 2 cases:

- 1st card is red joker or 2 card is a red ace
P1 = 1 - P(1st isn't red joker and 2nd isnt red ace)
= 1 - (53/54)(51/53)

- 1st card is red ace, or 2nd is a red joker
P2 = 1 - P2'
= 1 - (52/54)(52/53)

Then P = P1 + P2.
Well, I hope it will give you right answer :wink:

what about if you get 2 red aces?
Reply 10
Ah yes, forgot there can be 2 red aces.
So take
P3 = 1 - P3'
= 1 - (52/54)(52/53)

Erh...no wait ... it might be wrong if add them up as maybe it may overlap with other cases
Reply 11
Hmm, after thinking again :biggrin:, well, is that case is included in 2 other cases?
Cuz I did with "or"... ???
Reply 12
None of those worked for number 1. The right answer in my book is 52/477 unless its wrong in my book. Ohh are there any sites that explain how to do complimentary things like how you put the apostrophe in, because my teacher does not teach at all.

I have two more questions for now.

A test has four true/false. Determine:
a) The probability that a student will get all four correct by guessing.
b) The probability that a student will get 3 correct by guessing

My second question is this.
An airplane can make a safe landing if at least half of its engines are working. Suppose that engine failures are independant events. Determine whether a two-engine plane is safer that a four engine plane if the chance that an engine fails is 1 in 2.

Please help me with these last questions.
F.A.B.
None of those worked for number 1. The right answer in my book is 52/477 unless its wrong in my book.


my answer and solution above is correct then because
1- (51/54)*(50/53)= 1- 425/477 = 52/477
F.A.B.

A test has four true/false. Determine:
a) The probability that a student will get all four correct by guessing.
b) The probability that a student will get 3 correct by guessing


This is straightforward.

(a) P(correct)= 1/2, then P(4 correct)= (1/2)^4

(b) If you havent done factorials or combinations, then think of this in your head kinda like a tree diagram. If you want exactly 3 right answers, howmany different ways can you get 3 right answers?
You could:
get the 1st wrong, next 3 right
1st right, 2nd wrong, last 2 right
1st & 2nd right, 3rd wrong, last right
1st, 2nd & 3rd right, last wrong

So there are four possible different ways of getting exactly 3 right. You work out the probablity of either one above then multiply by 4 since there are four of them. I'll let you do that bit :biggrin:
F.A.B.

My second question is this.
An airplane can make a safe landing if at least half of its engines are working. Suppose that engine failures are independant events. Determine whether a two-engine plane is safer that a four engine plane if the chance that an engine fails is 1 in 2.


Well, work out the probablity, in a two-engine plane, of both engines failing (as this is the only way in which it wont make a safe landing).

Then work out the probablity of three or more engines failing in a four-engine plane. This is not as straightforward. You have to work out probablity of all four engines failing, then add the probablity of only three engines failing by listing the poss combinations as in the previous question.

Compare and see which one is more likely not to make a safe landing, and hence which is safer.
Reply 16
silent ninja how did you come up with the answer to number 1. You did this
1- (51/54)*(50/53)= 1- 425/477 = 52/477

Where did you get 51/54 and 50/53?

Please anyone i really need to know
Reply 17
I think what he did is:
- Find the probability of not taking 3 cards(2 red aces and 1 red joker) in 2 draws.
So he had 51/54 in first draw, and 50/53 in the 2nd draw.
F.A.B.
silent ninja how did you come up with the answer to number 1. You did this
1- (51/54)*(50/53)= 1- 425/477 = 52/477

Where did you get 51/54 and 50/53?

Please anyone i really need to know


Well since P(red joker)= 1/54, P(Red Ace)= 2/54
This leaves us with 51 cards in the pack that are neither of the above i.e. P(neither)= 51/54

If the cards are not replaced, then on your second turn there are only 53 cards remaining.

Since you require 1- P(neither)*P(neither) to give you the probablity of at least a red ace or red joker:
then on first go, P(neither)= 51/54
on second go, P(neither)= 50/53 --there is one less card in the pack and also one less 'neither' card remaining.

Hope this clears things up.