How do i solve this: A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die ??

solution(1)-probability for an event to occur=(no. of favourable outcomes/total no. of outcomes) in this case, probability of obtaining a six =(no. of cases in which we obtain a six/total no of cases) sample space S={1,2,3,4,5,6} P(E)=n(E)/n(S) P(6)=1/6 P(6^)(read as 6 bar)=1-P(6)=1-(1/6)=5/6 6th throw:we want to get only six ,so P(6) out of all remaining throws, in 2 throws we want 6 so P(6) in rest 3 throws we want anything except 6,so P(6^) P=P(6)*P(6^)*P(6^)*P(6)*P(6^)*P(6)=(1/6)*(5/6)*(5/6)*(1/6)*(5/6)(1/6)=125/46656=0.002679184 Solution(2)- third six must come in the sixth throw of a die i,e. two sixes must come in first 5 throw n=5,p=probability of getting a six=1/6,q=5/6 using binomial distribution P(r)=5cr*(5/6)power 5-r*(1/6)power r required probability=P(2)* 1/6=5c2 *(5/6)power 3 *(1/6) power 2 * (1/6)=0.0267 read less

Our situation is something like this. Since we stop after getting three sixes and since the third six must come in the sixth position, we have our situation as, X X X X X 6 The first five places are to be filled by two sixes in any combination. For such 6 combinations, as in X X X X X 1 X X X X X 2 X X X X X 3 X X X X X 4 X X X X X 5 X X X X X 6 .... only this last combination is relevant. Such a combination occurs only 1/6 of the total number of combinations. Now in the remaining five positions only two positions need to be fixed by two sixes and the rest of the 3 positions to be not sixes. If p = 1/6 is the probability of getting a six with one die then q = 1 - (1/6) = 5/6 is the probability of not getting a six. We can place two sixes in 5 positions in 5C2 ways and therefore the two sixes will have a probability of 5C2 * (1/6)*(1/6)*(5/6)*(5/6)*(5/6) = [ 10 * (5^3) / (6^5) ] Now we only need one sixth of this result as explained before. Hence the required probability is [ 10 * (5^3) / (6^5) ] / 6 = 1250 / (6^6) Hope this helps, Cheers read less

the probability of obtaining the third six in the sixth throw of the die = the probability of obtaining two sixes in the first five throws and the third six in the sixth throw of the die P(6) = 1/6 and P(not 6) = 1-(1/6) = 5/6 Therefore,, p=1/6 and q=5/6 and n=5 P(X=r)=5Crprq(5-r) ----------------(1) Put the value of p,q and r=2 into equation (1) and after calculation we got.. P(X=r) = 1250/6*6*6*6*6 Ans.... Here i Can't able to write in the form for Permutation and combination...so message me in my inbox and you can join my free online class for this particular questions....I will show you how to solve this question..... Good Luck,,,,,, read less

This is a problem of binomial distribution with p = 1/6. The condition stated in the question is that you should get the third 6 in the sixth throw. That means you should get exactly two 6's in the first 5 throws. This probablity is 5C2 x (1/6)^2 x (5/6)^3. This should be multiplies by the probability of getting a 6 in the last throw, which is 1/6. So the answer is 0.0268. read less

Dear Pratik, Ignore my previous answer; I made two mistakes (a) I did not read your question well enough (was in a hurry being occupied in mind with some other work) (b) I employed occurrence of a few number of sixes in the successive throws of dice having the number of possible ways according to possible permutations where as each occurrence of number six being identical, it must be combinations instead of permutations. With due apologies, I sent the right answer while trying to make the concept/logic clear enough: Question: How do i solve this: A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die ?? Class XI-XII Tuition (PUC) Answer: The logic to the solution comes from the fact P(six) = Product of [ P’(six),(P’’(six)) and (P (Numbers other than 6]) [equation (1)] where, (i) P(six) =The probability of occurrence of sixes three times, having 3rd six occurring at sixth throw (ii) P’(six) = Probability of occurrence of number six at the sixth throw = (1/6) (iii) P’’(six)=(means the probability of occurrence of sixes two times for five throws of dice with the numbers occurring for the remaining three throws {x,y,z} fixed and (iv) P (Numbers other than 6) = The probability of occurrence of numbers other than six (ie occurrence of any of the five elements out of {1,2,3,4,5} for {x,y,z} during the remaining three throws. (a) Now, probability of occurrence of a the number 6 out of {1,2,3,4,5,6} for a sixth throw of dice = P’(six) = ( 1/6) (b) With {x,y,z} fixed the possible combinations two sixes can occur in first five throws of the dice = 5C2 (note: here C stands for combination) Therefore P’’(six) = 5C2*(1/6)*(1/6) [because the probability for occurrence of number six for one throw is (1/6)] =[(5!)/(3!*2!*6*6)=(5/18) (c) Possible ways the numbers other than 6 out of {1,2,3,4,5,6} can occur for three throws of dice =5*5*5 [for every number in z there are five possible numbers for y and for every possible yz there are 5 possible numbers for x] For occurrence of each number in xyz the probability is (1/6) Therefore, P (Numbers other than 6) = (5*5*5)*(1/6)*(1/6)*(1/6) 2. With {x,y,z} taking any values out of the five numbers {1,2,3,4,5), we see that This means P(six) = Product of [ P’(six),(P’’(six)) and (P (Numbers other than 6)] = (1/6)*(5/18)*(125/216) =(625/23328) With regards, K Rajagopalan Nair read less