~Probility~

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...........

:)

28 Answers

13
Двҥїяuρ now in medical c ·

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

13
Двҥїяuρ now in medical c ·

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*0.44*0.63+7C5*0.45*0.62+7C6*0.46*0.61+7C7*0.47=4528/15625=.289792

23
Abhishek Priyam ·

:)

23
Abhishek Priyam ·

SOS

23
Abhishek Priyam ·

Yeah Abhirup thats .2897 or what u wrote but how ????/

11
Anirudh Narayanan ·

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]

23
Abhishek Priyam ·

Language?

11
Anirudh Narayanan ·

5. 1/4 [4]

23
Abhishek Priyam ·

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

23
Abhishek Priyam ·

Cheers 5) 1/4 me too but i think answer given is wrong

11
Anirudh Narayanan ·

Printing mistake

11
Anirudh Narayanan ·

Σ
γνώμη φίλε. Δεν μπορούν να λύσο
ν τα
πόλοιπα. [2]

23
Abhishek Priyam ·

Aragorn: Sorry buddy. Can not solve the other :(

That was Greek.. :P

341
Hari Shankar ·

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 7C4 (0.4)4(0.6)3.

1
Mayukh Bagchi ·

Can anyone show how to do the second question.

23
Abhishek Priyam ·

Anyone......

[7]

23
Abhishek Priyam ·

Yup that was written in question which i missed here to write... but i think Abhirup knows everyones man ki baat :P

1
Honey Arora ·

Q2) d
coz i think n=128

Q3)521216/96, using bernaulli's trials

13
Двҥїяuρ now in medical c ·

m i right in ques no. 2 ????...is b) the right ans???

23
Abhishek Priyam ·

Sorry it is 3/32 :(

13
Двҥїяuρ now in medical c ·

oops ...it is 48/512

there are 48 such cases

23
Abhishek Priyam ·

How?? [7] method plz....

13
Двҥїяuρ now in medical c ·

* 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

23
Abhishek Priyam ·

So counting.... hmmmmmmmm
I dind't wanted to count.. :P
Is there other method??

any other method anyone.......

13
Двҥїяuρ now in medical c ·

.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

13
Двҥїяuρ now in medical c ·

no other method cmin in my mind....:(
hope there is a better method....keep trying

23
Abhishek Priyam ·

Now its 3 memeber for 1/4 cheers...

13
Двҥїяuρ now in medical c ·

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.......@!#@!%,....:(

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