Problem Solving

N:Dure wrote:1) A coin is tossed 6 times. What's the probability of getting at least 4 heads on consecutive tosses?

a) 1/64
b) 3/32
c) 15/64
d) 11/32
e) 90/64


2) Given a positive integer K

A) The probability that (K) (K+1) (K+2) (K+3) (K+4) is divisible by 20
B) 1

a) A is bigger
b) B is bigger
c) both are equal
d) Insufficient


3) Given that 'n' numbers are 8 i.e. (8,8,8,8,8....n) and 'n+2' numbers are 10. What's the arithmetic mean for both sets?

N:Dure wrote:Namaste! Thanks Anush for your help.

Just to reiterate, in no.2 you assume K=1 so 2,3, 4, 5 making them multiples and 20 is 2*2*5 so they have common factors, and the product of Ks is also / 20, right?
I also thought the same for no.3 but the answer is (9n+10)/(9n+1) don't ask me how

N:Dure wrote:Another q: If 65% of the total stamps are foreign, and 35% of the stamps with last ten years (?), and 40% are the foreign stamps ten years ago (?). If the total stamps are 1200, then what number of stamps are ten years now?

beat_gmat_09 wrote:Hi,
The question asks for at least 4 heads.
1st we get 4 heads remaining 2 has to be Tails. This can happen - HHHHTT, number of ways heads can occur = 6!/(2!*4!) = 15
2nd we get 5 heads remaining 1 has to be tail. This can happen - HHHHHT, number of ways heads can occur = 6!/5! = 6
3rd we get all 6 heads. This can only happen in one way i.e. HHHHHH
Probability of getting heads or tails = 1/2, occurance is 6 times, thus probability of occurance either heads or tails - (1/2)^6
We need to combine the above cases, treating them as 'OR' with the probability (1/2)^6 multiplied by the number of times they can occur.

Thus required probability -
= P(4 Heads,2 tails) + P(5 heads 1 tail) + P(all 6 heads)

=[15*(1/2)^6] + [6*(1/2)^6] + [(1/2)^6]

Take out (1/2)^6 common term

=(1/2)^6[15+6+1]
= 22/64
= 11/32
HTH

1st we get 4 heads remaining 2 has to be Tails. This can happen - HHHHTT, number of ways heads can occur = 6!/(2!*4!) = 15

anshumishra wrote: The question asks for at least 4 heads on consecutive tosses.

1st we get 4 heads remaining 2 has to be Tails. This can happen - HHHHTT, number of ways heads can occur = 6!/(2!*4!) = 15

This will contain some ways such as (HTHTHH, THHHTH, etc...) which doesn't satisfy the criteria.

Correct me if I am wrong, But that is what the question meant to me. It is not asking the probability of just getting at least 4 heads (but rather at least 4 consecutive heads : HHHHTT, THHHHT,TTHHHH,HHHHHT,THHHHH,HHHHHH).

at least 4 heads on consecutive tosses

beat_gmat_09 wrote:

anshumishra wrote: The question asks for at least 4 heads on consecutive tosses.

1st we get 4 heads remaining 2 has to be Tails. This can happen - HHHHTT, number of ways heads can occur = 6!/(2!*4!) = 15

This will contain some ways such as (HTHTHH, THHHTH, etc...) which doesn't satisfy the criteria.

Correct me if I am wrong, But that is what the question meant to me. It is not asking the probability of just getting at least 4 heads (but rather at least 4 consecutive heads : HHHHTT, THHHHT,TTHHHH,HHHHHT,THHHHH,HHHHHH).

at least 4 heads on consecutive tosses

The tosses are consecutive, not the heads getting on the tosses !

anshumishra wrote: How does toss being consecutive adds anything to the question ? What is the difference between these two questions :

A coin is tossed 6 times. What's the probability of getting at least 4 heads on consecutive tosses?
vs
A coin is tossed 6 times. What's the probability of getting at least 4 heads ?

beat_gmat_09 wrote:

anshumishra wrote: How does toss being consecutive adds anything to the question ? What is the difference between these two questions :

A coin is tossed 6 times. What's the probability of getting at least 4 heads on consecutive tosses?
vs
A coin is tossed 6 times. What's the probability of getting at least 4 heads ?

As far as my knowledge both are same.
The consecutive notion which you are talking about requires the question to state explicitly that the tosses on the coins should be heads consecutively i.e. HHHHTT, THHHHT .....
I think the problem should state something like this - ... of getting at least 4 heads consecutively on the tosses (the tosses are consecutive here)