Probability Questions (1 Viewer)

Skeptyks

Member
Joined
May 6, 2011
Messages
395
Gender
Male
HSC
2012
I'm going to update this thread with probability questions that I cannot do as I go through like 6 exercises of them.
Any help is appreciated, thanks in advance :]

1) From an ordinary deck of cards, only 12 picture cards are retained. They are shuffled and a man draws 2 cards at random and announces that he holds at least one King. Find the probability that he holds two kings in his hand.

2) A four letter 'word' is made out of the letters from REASON. If R and N are included how many words can be made?
I did (4 x 3) x 4 x 3 which gave the correct answer but I don't remember how I did it. Could anyone please explain?


3) The letters of the word 'integral' are used to make a word. If 3 consonants and 2 vowels are selected from the letters of the word 'integral' What is the probability that the letters T, E are chosen? Explanation brief/long also if you can please ^^


Figured out last one its all good. Thanks for all your help guys.
 
Last edited:

bleakarcher

Active Member
Joined
Jul 8, 2011
Messages
1,509
Gender
Male
HSC
2013
For the second one you realised that when arranging R you have 4 places to put it and that after you arrange R you have 3 remaining places for N. There remain 2 more letters to form the four letter 'word' from the 4 remaining. Hence, the number of ways you can make the word from the letters of the word 'REASON' if it must contain R and N is 4*3*4*3.
 

bleakarcher

Active Member
Joined
Jul 8, 2011
Messages
1,509
Gender
Male
HSC
2013
From an ordinary deck of cards, only 12 picture cards are retained. They are shuffled and a man draws 2 cards at random and announces that he holds at least one King. Find the probability that he holds two kings in his hand.

In the first problem, the man announces at least one of the two cards is a King from the four Kings in the deck. If the second card is to be a King as well, there are three more Kings to choose from. Hence, the number of ways the two Kings can be drawn is 3*4. However, the question asks for probability. After the first King is chosen there are 51 cards left. Hence, the number of possible ways the two cards can be drawn becomes 51*4. Hence, the probability of picking two Kings becomes (3*4)/(51*4)=3/51=1/17.
 

bleakarcher

Active Member
Joined
Jul 8, 2011
Messages
1,509
Gender
Male
HSC
2013
I did that based on logic, I dont have too much experience with these questions.
 

Skeptyks

Member
Joined
May 6, 2011
Messages
395
Gender
Male
HSC
2012
Sorry there is actually on 12 cards. The other 40 cards in the deck are thrown away.

*talking to bleakarcher.
 
Last edited:

Skeptyks

Member
Joined
May 6, 2011
Messages
395
Gender
Male
HSC
2012
Your first question is a different take on a notorious HSC question from 1997. This question says "A bag contains 2Red, 1 white and 1 blue ball. Two ball are randomly chosen, and one of these is randomly dropped on the floor. If the ball on the floor is red, what is the probability that the ball still n your hand is also red".

The answer that was marked correct at the time was 1/5. The correct answer (it was officially changed 10 years later!) is 1/3. The answer of 1/5 IS the correct answer to a slightly different question: "After choosing the two balls, I announce that at least one of the balls is red. What is the probability that they are both red". Your question is of the same type as this second question.

The number of possible 2-card hands is 12C2 = 66. The number of possible 2-card hands NOT containing a King is 8C2=28. So the number of 2-card hands containing AT LEAST ONE KING is 66-28=38.
The number of 2-card hands containing 2 Kings is 4C2=6.

So the Probability of having two Kings GIVEN THAT YOU HAVE AT LEAST ONE KING is 6/38=3/19.

**Lets modify the question so that it is like the HSC question instead: "A hand you a card and you see that it is a King. What is the probability that the other card is also a King?
In that case, the answer would have been: There is one King left from 11 cards, so 1/11.

Why are the answers not the same? Because of the different amount of information used to describe the scenario. In one case, you are only told that one of the two cards is a King. It the other case, you are told exactly WHICH card is a King.


NB. I am a maths teacher. The previous answer is incorrect.
Brilliant explanation, thanks for that! :p I will be posting some others I can't do in a few minutes/hours. Thankyou again for spending your time to do that haha.
 

Skeptyks

Member
Joined
May 6, 2011
Messages
395
Gender
Male
HSC
2012
From an ordinary deck of cards, only 12 picture cards are retained. They are shuffled and a man draws 2 cards at random and announces that he holds at least one King. Find the probability that he holds two kings in his hand.

In the first problem, the man announces at least one of the two cards is a King from the four Kings in the deck. If the second card is to be a King as well, there are three more Kings to choose from. Hence, the number of ways the two Kings can be drawn is 3*4. However, the question asks for probability. After the first King is chosen there are 51 cards left. Hence, the number of possible ways the two cards can be drawn becomes 51*4. Hence, the probability of picking two Kings becomes (3*4)/(51*4)=3/51=1/17.
Thanks for answering the 1st question! :p I can't believe that I didn't remember how to do that...
 

bleakarcher

Active Member
Joined
Jul 8, 2011
Messages
1,509
Gender
Male
HSC
2013
Sorry didnt read the first line lol. Thought all 52 cards were used.
 

Skeptyks

Member
Joined
May 6, 2011
Messages
395
Gender
Male
HSC
2012
Is it important to write detailed reasons and explanations to answers in probability questions? Or is it fine to restate the question and proceed with the answer??
 

Skeptyks

Member
Joined
May 6, 2011
Messages
395
Gender
Male
HSC
2012
Oh okay. So just if it is a probability question write P(whatever it is) = calculations.
For a question that asks for total number of combinations I would just write, Total number of combinations = calculations.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top