i go for C 1/2raunekk wrote:Q)A drawer holds 4 red hats and 4 blue hats.What is the probability of getting exactly 3 red hats or exactly 3 blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?
(a)1/8
(b)1/4
(c)1/2
(d)3/8
(e)7/12
Your mistake is here. It is 2^4 = 16 for 4 positionsraunekk wrote: Total number of ways in selectin 4 balls is 8C4 ways..
Stuart,Stuart Kovinsky wrote:Any time there's a 50/50 chance of something happening, you really have a coin flip question in disguise. The question could have been:
If you flip a fair coin 4 times, what's the probability of getting exactly 3 heads or exactly 3 tails?
There are numerous ways to solve coin flip questions. One of the quickest is to apply the coin flip formula.
The probability of getting exactly k results out of n flips is:
nCk/2^n
Applying the formula to this question, we get:
4C3/2^4 = 4/16 = 1/4
Since we want exactly 3 heads OR exactly 3 tails, we need to double our answer, getting 1/2.
4C3 in the coin flip question does account for all 4 flips - we want exactly 3 of them to be heads, by default the other will be tails.Stockmoose16 wrote:Stuart,Stuart Kovinsky wrote:Any time there's a 50/50 chance of something happening, you really have a coin flip question in disguise. The question could have been:
If you flip a fair coin 4 times, what's the probability of getting exactly 3 heads or exactly 3 tails?
There are numerous ways to solve coin flip questions. One of the quickest is to apply the coin flip formula.
The probability of getting exactly k results out of n flips is:
nCk/2^n
Applying the formula to this question, we get:
4C3/2^4 = 4/16 = 1/4
Since we want exactly 3 heads OR exactly 3 tails, we need to double our answer, getting 1/2.
I don't understand your logic here. In the numerator, don't you have to account for the fourth flip? Shouldn't the numerator be 4C3* 4C1?
Look at the following example:
Three letters are randomly chosen from the word TUESDAY. How many possible selections are there? How many of these selections have:
(b) exactly 2 vowels?
No. of selections = 3C2* 4C1= 12
How come for the coin flip question, you don't have to do:
4C3 * 4C1 for the numerator?
Got it. Thanks for the explanation. It's very helpful.4C3 in the coin flip question does account for all 4 flips - we want exactly 3 of them to be heads, by default the other will be tails.
In the "TUESDAY" question, the "3" in 3C2 doesn't relate to the number of letters we're choosing, it relates to the number of vowels available (we have 3 vowels total, we're choosing 2 of them); similarly, the "4" in 4C1 relates to the number of consonants.
So, we have 3C2*4C1 because we're choosing 2 vowels out of 3 and 1 consonant out of 4.
In our example, we have 4C3 because we're choosing 3 results out of 4 flips - we're not choosing 3 different "heads" out of a total of 4 available heads.
Stuart,Stuart Kovinsky wrote:4C3 in the coin flip question does account for all 4 flips - we want exactly 3 of them to be heads, by default the other will be tails.Stockmoose16 wrote:Stuart,Stuart Kovinsky wrote:Any time there's a 50/50 chance of something happening, you really have a coin flip question in disguise. The question could have been:
If you flip a fair coin 4 times, what's the probability of getting exactly 3 heads or exactly 3 tails?
There are numerous ways to solve coin flip questions. One of the quickest is to apply the coin flip formula.
The probability of getting exactly k results out of n flips is:
nCk/2^n
Applying the formula to this question, we get:
4C3/2^4 = 4/16 = 1/4
Since we want exactly 3 heads OR exactly 3 tails, we need to double our answer, getting 1/2.
I don't understand your logic here. In the numerator, don't you have to account for the fourth flip? Shouldn't the numerator be 4C3* 4C1?
Look at the following example:
Three letters are randomly chosen from the word TUESDAY. How many possible selections are there? How many of these selections have:
(b) exactly 2 vowels?
No. of selections = 3C2* 4C1= 12
How come for the coin flip question, you don't have to do:
4C3 * 4C1 for the numerator?
In the "TUESDAY" question, the "3" in 3C2 doesn't relate to the number of letters we're choosing, it relates to the number of vowels available (we have 3 vowels total, we're choosing 2 of them); similarly, the "4" in 4C1 relates to the number of consonants.
So, we have 3C2*4C1 because we're choosing 2 vowels out of 3 and 1 consonant out of 4.
In our example, we have 4C3 because we're choosing 3 results out of 4 flips - we're not choosing 3 different "heads" out of a total of 4 available heads.
You're misunderstanding how to apply the combinations formula. To be honest, I have no clue how you derived the expression above.Stockmoose16 wrote:I actually have one follow-up question. In regards to this question:
A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?
(a) 1/8
(b) ¼
(c) ½
(d) 3/8
(e) 7/12
I understand you treat this like a coin flip, but if you didn't recognize that from the beginning, why wouldn't this be exactly like the vowel/consonant question I was referred to above. Meaning, you should multiply the ways of getting exactly 3 red hats times the probability of getting a blue one on the fourth choice? Then multiply by two to account for the "or."
So the answer should be:
4C3*4/2^4 +4C3*4/2^4 =2 ... which obviously makes no sense. Why is this question different from the vowel one above?
