1. Prob of A wining a game is 0.4 while that of B is 0.6 . What is the probability of A wining a best of 7 games?
7C4*0.4^{4}*0.6^{3}+7C5*0.4^{5}*0.6^{2}+7C6*0.4^{6}*0.6^{1}+7C7*0.4^{7}=4528/15625=.289792
Plz help me in these questions..............(FIITJEE Package)
1. Prob of A wining a game is 0.4 while that of B is 0.6 . What is the probability of A wining a best of 7 games?
2. If Ram and Shyam selects two numbers at random with replacements from {1,2,3,4.....n}. If the probability that Shyam selects a number less than one selected by Ram is 63/128 then...
a) n is odd b) n is perfect square
c) n is a perfect 4th power d) None
3. The probability that a 6 digit number selected at random contains at most three digits repeated is............
4. A fair coin is tossed 9 times the probability that at least 5 consecutive heads occurs is............
5.Let A B C be 3 independent events such that P(A)=1/3, P(B)=1/2, P(C)=1/4. Then probability that exactly 2 events occurs out of 3 events is...........
:)
2. If Ram and Shyam selects two numbers at random with replacements from {1,2,3,4.....n}. If the probability that Shyam selects a number less than one selected by Ram is 63/128 then...
a) n is odd b) n is perfect square
c) n is a perfect 4th power d) None
my ans is(b)
i m getting n=64
1. Prob of A wining a game is 0.4 while that of B is 0.6 . What is the probability of A wining a best of 7 games?
7C4*0.4^{4}*0.6^{3}+7C5*0.4^{5}*0.6^{2}+7C6*0.4^{6}*0.6^{1}+7C7*0.4^{7}=4528/15625=.289792
1.Binomial distribution problem. Consider the case where R = 4 or 5 or 6 or 7. Only then can win the tournament, or whatever it is.[4]
Aragorn:
Binomial distribution problem. Consider the case where R = 4 or 5 or 6 or 7. Only then can win the tournament, or whatever it is
Aragorn: Sorry buddy. Can not solve the other :(
That was Greek.. :P
For Abhirup:
In the 1st question, suppose we take the case where A wins 4 games to take the tourney
Now i he wins 4 games on a trot, does he still need to play 3 more games? That is the assumption when you write ^{7}C_{4} (0.4)^{4}(0.6)^{3}.
Yup that was written in question which i missed here to write... but i think Abhirup knows everyones man ki baat :P
* denote anyone of heads or tails.....
HHHHHT***....8
HHHHHHT**....4
HHHHHHHT*....2
HHHHHHHHT.....1
HHHHHHHHH....1
THHHHHT**...4
THHHHHHT*..2
THHHHHHHT..1
THHHHHHHH...1
*THHHHHT*...4
*THHHHHHT...2
*THHHHHHH...2
**THHHHHT...4
**THHHHHH...4
***THHHHH.....8
TOTAL CASES 48
PROBABILITY......48/512
So counting.... hmmmmmmmm
I dind't wanted to count.. :P
Is there other method??
any other method anyone.......
.Let A B C be 3 independent events such that P(A)=1/3, P(B)=1/2, P(C)=1/4. Then probability that exactly 2 events occurs out of 3 events is...........
1/3*1/2*3/4=1/8
1/2*1/4*2/3=1/12
1/4*1/3*1/2=1/24
adding...1/8+1/12+1/24=1/4
no other method cmin in my mind....:(
hope there is a better method....keep trying
1. Prob of A wining a game is 0.4 while that of B is 0.6 . What is the probability of A wining a best of 7 games?
is it 4528/15625.......@!#@!%,....:(