@kleptomaniac_20 said:@gnehagarg : A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.a) 3/4b) 3/8c) 1/2d) 2/3Can you please explain ?
3/8?
@sauravd2001 said:Let X be a four-digit positive integer such that the unit digit of X is prime and the product of all digits of X is also prime. How many such integers are possible?
i'll say there are pnly four such numbers:
1112
1113
1115
1117
@kleptomaniac_20 said:@gnehagarg : A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.a) 3/4b) 3/8c) 1/2d) 2/3Can you please explain ?
Should be 3/4..Proved yesterday by @chillfactor Sir
P.S. I know the answer given is 3/8. NCERT Solved example
Check it out: http://www.pagalguy.com/posts/4128090
@kleptomaniac_20 said:@gnehagarg : A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.a) 3/4b) 3/8c) 1/2d) 2/3Can you please explain ?
prob of speaking truth = 3/4
prob of speaking false = 1/4
so 3/4*1/6 / 3/4*1/6 + 1/4*5/6 = 3/8
@kleptomaniac_20 said:@gnehagarg : A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.a) 3/4b) 3/8c) 1/2d) 2/3Can you please explain ?
P(speaks truth)=3/4
P(lies)=1/4
P(actually a six)=P(getting a six on the die)P(he reports it correctly)/same+P(6 on the die)P(reports wrongly)
= 1/6*3/4/(1/6*3/4+1/6*1/4)
=1/8/1/6
=3/4
where am i going wrong???
@kleptomaniac_20 said:@19rsb Approach please !!
P(speaking truth)=3/4....P(false)=1/4
p(six on dice)=1/6...........P(other than 6)=5/6
Conditional probability=p(truth)*p(six)/ p(truth)*p(six)+p(false)*p(not six)
=(3/4)*(1/6)/(3/4)*(1/6)+(1/4)*(5/6)
=3/8
p(six on dice)=1/6...........P(other than 6)=5/6
Conditional probability=p(truth)*p(six)/ p(truth)*p(six)+p(false)*p(not six)
=(3/4)*(1/6)/(3/4)*(1/6)+(1/4)*(5/6)
=3/8
@vijay_chandola said:Let N be a 4 digit number in base 5. A converts it into decimal and finds that it is divisible by 7. M on the other hand assumes it to be a decimal number and observes that it is also divisible by 7. How many such N exist?
If 'abcd' is the 4 digit number in base 5, then
125a + 25b + 5c + d is divisible by 7
=> 4b + d - a - 2c is divisible by 7
Also, 1000a + 100b + 10c + d is divisible by 7
=> 2b + d - a + 3c is divisible by 7
Using these two conditions we can say that (4b + d - a - 2c) - (2b + d - a + 3c) = 2b - 5c is divisible by 7
Possible values of (b, c) are (3, 4), (4, 3), (0, 0)
When (b, c) = (3, 4)
4b + d - a - 2c is divisible by 7
=> d - a + 4 is divisible by 7
=> (a, d) = (4, 0), (1, 4)
When (b, c) = (4, 3)
4b + d - a - 2c is divisible by 7
=> d - a + 10 is divisible by 7
=> (a, d) = (4, 1), (3, 0)
When (b, c) = (0, 0)
4b + d - a - 2c is divisible by 7
=> d - a is divisible by 7
(d, a) = (1, 1), (2, 2), (3, 3), (4, 4)
So, only 8 such numbers are there 4340, 1344, 4431, 3430, 1001, 2002, 3003, 4004
Solved example of NCERT:- Check page no 555 example 21
But I got 3/4 as answer:-
@kejriwalsonal1 : Sir he has reported that he has got a six, then why are you considering the case when he gets a six and reports it wrongly ?
@kejriwalsonal1 said:P(speaks truth)=3/4 P(lies)=1/4P(actually a six)=P(getting a six on the die)P(he reports it correctly)/same+P(6 on the die)P(reports wrongly) = 1/6*3/4/(1/6*3/4+1/6*1/4) =1/8/1/6 =3/4where am i going wrong???
Bold part should be P(6 didn't appear on the dice). In your case if 6 appears and he lied, means he will definitely report something different from 6
Can anybody post any link for conditional probability lesson ? Need to brush up my concepts !!
Thanks !!
@19rsb said:P(speaking truth)=3/4....P(false)=1/4p(six on dice)=1/6...........P(other than 6)=5/6Conditional probability=p(truth)*p(six)/ p(truth)*p(six)+p(false)*p(not six)=(3/4)*(1/6)/(3/4)*(1/6)+(1/4)*(5/6)=3/8
In this case, how did you take 1/4*5/6...?
it includes the cases when he is lying that let say, 5 has come. Does that necessarily means 6 would come...?
@vijay_chandola said:In this case, how did you take 1/4*5/6...?it includes the cases when he is lying that let say, 5 has come. Does that necessarily means 6 would come...?
sir ji according to my understanding there will be 2 case either 6 will come or 6will not come(any no other than six)
now question is to find prob when it is actually a six
then CASE 1.....it is a six and he reported it truly
CASE 2 It is not a six(other no) and he reports a six(lying)
means he reports a six in either cases but we have to find prob when it is actually a 6.......and rest is the conditional prob concept......
I 'm not sure as I had done it correctly or not becoz 2 or 3 ppl abuv are getting 3/4
@chillfactor sir where to go either with 3/4 or with 3/8
now question is to find prob when it is actually a six
then CASE 1.....it is a six and he reported it truly
CASE 2 It is not a six(other no) and he reports a six(lying)
means he reports a six in either cases but we have to find prob when it is actually a 6.......and rest is the conditional prob concept......
I 'm not sure as I had done it correctly or not becoz 2 or 3 ppl abuv are getting 3/4
@chillfactor sir where to go either with 3/4 or with 3/8
@19rsb said:sir ji according to my understanding there will be 2 case either 6 will come or 6will not come(any no other than six)now question is to find prob when it is actually a sixthen CASE 1.....it is a six and he reported it trulyCASE 2 It is not a six(other no) and he reports a six(lying)means he reports a six in either cases but we have to find prob when it is actually a 6.......and rest is the conditional prob concept......I 'm not sure as I had done it correctly or not becoz 2 or 3 ppl abuv are getting 3/4@chillfactor sir where to go either with 3/4 or with 3/8
I took the same cases as you,
Case 1:- It is a 6 and he reported a 6
(1/6)(3/4) = 3/24
Case 2:- It is not a 6 and he reported a 6 (he lied)
When he lies, he has 5 different possibilities to lie (suppose 5 appeared on the die, then he can say 1 or 2 or 3 or 4 or 6, so probability that he will say 6 will be 1/5)
so, probability = (5/6)(1/4)(1/5) = 1/24
So, required probability = (3/24)/{(3/24) + (1/24)} = 3/4
@chillfactor said:I took the same cases as you,Case 1:- It is a 6 and he reported a 6(1/6)(3/4) = 3/24Case 2:- It is not a 6 and he reported a 6 (he lied)When he lies, he has 5 different possibilities to lie (suppose 5 appeared on the die, then he can say 1 or 2 or 3 or 4 or 6, so probability that he will say 6 will be 1/5)so, probability = (5/6)(1/4)(1/5) = 1/24So, required probability = (3/24)/{(3/24) + (1/24)} = 3/4
I wanted to share my point of view regarding this problem, (you may or may not agree).
What I have observed with all the solutions presented till now on the forum is that we try to apply the concept of conditional probability.
Now, as per my understanding we apply conditional probability when we need to re-asses the probability in a given scenario under the light of a new information provided.
The key thing we should assess before we recalculate the probabilities is to ask ourselves that are there any cases which I was considering earlier is getting ruled out under the light of new information?
for example in the question it says that p(truth) = 3/4. Now over here his probability of speaking the truth is not a function of what number it appears on the dice, so irrespective of whatever number appears on the dice his probability of speaking the truth remains unchanged. So even if we know the final outcome our initial estimates of probabilities wont change. so it has to be (3/4)
Had the question been that p(truth) = 3/4 if odd number appears on dice, and p(truth)= 3/5 if even number appears on the dice. In this case we know that his speaking of truth is function of what appears on dice, so in this case the probability needs to be re-assessed.
Another Example
To illustrate my point here is an example. (My favorite example for conditional prob.)
say you fall from sky after an airplane accident. There is (1/3) probability that you fall on land and (2/3) probability that you fall on water. In the process of falling you went unconscious. It was after 1 month you come out coma and realize that you are still alive. Now you are wondering what is the probability that you fell in water?
Case1: if the probability of surviving by falling on land was zero. Then you would know for sure that you had fallen on water (or else you would have been dead). So in this case the conditional probability of falling in water is 1.
Case2: if probability of surviving by falling on land was same as that by falling in water. In this case your knowledge that you finally survived is of no use, as it does not change your earlier estimates of falling in land or water.
Back to Question
So, I without applying any conditional probability my take is (3/4) as answer.
ATDH.
@kleptomaniac_20 said:@gnehagarg : A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.a) 3/4b) 3/8c) 1/2d) 2/3Can you please explain ?
A little late, but here's my take in brief:
There are two very similar problems we could have here, with small changes in wording but different answers.
One is the one stated above, where we would get 3/4 - "six and true" is 3/24, "not six and false" is 5/24 and in 15th of the latter cases i.e. 1/24 he will say 6. So it is actually 6 is 3/24 out of (3/24 + 1/24) = 3/4 cases.
(Details may be found in the excellent methods which several people have provided above, I shall not re-solve - I am only posting to clear up a trick in wording which is often encountered and which may be why someone said the OA was different).
The second way of phrasing it would be worded something like "A man is known to speak truth 3 out of 4 times. He throws a die, and when asked if it shows a six he says 'yes'. Find the probability that it is actually a six." For this question it would be 3/8. Basically here if it is not 6, a lie would be saying Yes (unlike in the other case when a lie would be saying a number other than the correct one, which would become 6 only 1/5th of the time.) So here probability "six and Yes" is 3/24, probability "Not 6 and Yes" = 5/24, so required probability is 3/(3+5) = 3/8.
Can someone check on the precise wording in the NCERT textbook? If it is the latter phrasing, then their answer is fine...if not then 3/4 hona chahiye.
regards
scrabbler
There are two very similar problems we could have here, with small changes in wording but different answers.
One is the one stated above, where we would get 3/4 - "six and true" is 3/24, "not six and false" is 5/24 and in 15th of the latter cases i.e. 1/24 he will say 6. So it is actually 6 is 3/24 out of (3/24 + 1/24) = 3/4 cases.
(Details may be found in the excellent methods which several people have provided above, I shall not re-solve - I am only posting to clear up a trick in wording which is often encountered and which may be why someone said the OA was different).
The second way of phrasing it would be worded something like "A man is known to speak truth 3 out of 4 times. He throws a die, and when asked if it shows a six he says 'yes'. Find the probability that it is actually a six." For this question it would be 3/8. Basically here if it is not 6, a lie would be saying Yes (unlike in the other case when a lie would be saying a number other than the correct one, which would become 6 only 1/5th of the time.) So here probability "six and Yes" is 3/24, probability "Not 6 and Yes" = 5/24, so required probability is 3/(3+5) = 3/8.
Can someone check on the precise wording in the NCERT textbook? If it is the latter phrasing, then their answer is fine...if not then 3/4 hona chahiye.
regards
scrabbler
Find all positive integers n for which n ˛ + 89n + 2010 is a perfect square.
@chillfactor said:Find all positive integers n for which n ˛ + 89n + 2010 is a perfect square.
Sir 0(no +ve integer) hai kya? 😞
@chillfactor said:Find all positive integers n for which n ˛ + 89n + 2010 is a perfect square.
sir ye factorize hua hai....(n+44)(n+45)+1
iske age kuch clear nai keh sakte...to is it 0??
but haan -44 aur -45 pe zrur hoga perfect square.