Can anyone explain this question to me!?

Question

Two fair coins are tossed together 172times. How many times would you expect to get:
a) twi heads,
b) a head and a tail in any order
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Thank you!!!!!!!!!!!!

Answer 1

Sorry for delay.


However coming to the point,
As you said the coins are fair. The probability of getting a head at the throw of each coin is 0.5 Also successive trials are indepedent. So the conditions of a bonomial distribution is fulfilled. The probailities of 0 head, 1 head, 2 heads are given by the successive terms in the expansion of the binomial expression,

(1/2 + 1/2)^2 = (1/2)^2 + nC1(1/2)(1/2) + (1/2)^2 ( Intentionally expanded using binomial theorem)

hence probality of getting two heads in a single trial( tossing the coins 1 time) = (1/2)^2= 1/4

Hence the expected no. of two heads in 172 trials= 172*1/4 = 43 ( ans of a )

Similarly the probabil;ity of getting 1 Head (in whatever order) in a single trial= 2C1(1/2)(1/2)= 1/2

Hence the expected no. of one head in 172 trials= 172*1/2 = 86 ( ans of b )

Answer 2

when you throw to coins , you have 4 possiblities

a) first head second head

b) first head, second tail

c) first tail, second head

d) first tail second tail

obtaining two heads only with case a) probability 1/4 . So with 172
, the probability of a is 172/4 = 43

Obtaining a head and a tail is case b) and c) . The probability is 1/2
If you try 172 , you theoretically obtain 172/2 =86

Answer 3

what a good question for a Sunday morning, just what I need!!

so the coins can fall

head head
tail tail
head tail
tail head

so 50% of the time you have a missmatch and 50% you have a match - and of that match only 50% of those will be head head .. so 25% of the throws should be head head and 25% of 172 is 43

50% of the throws is a mismatch - so 50% of 172 = 86

Answer 4

All possible combinations are:
HH
TH
HT
TT
It's a quarter chance for each as these are all the possibilities, probability adds up to 1, and there are 4 combinations, likelihood of getting any of them, as there is an equal chance, is therefore 1/4, a quarter.
So for two heads, a quarter of 172 is 43. That's your answer to a).
For a head and a tail in any order, that's 43 chanches for HT and 43 chances for TH, so overall that's 86. That's your answer to b).

Answer 5

a) two heads =1/4*172=43 times
b)a head and a tail in any order
=1/2*172=86 times

i hope that this helps

Answer 6

TAKE ANY SMALL FACTOR OF 172 eg- 4
hence if we toss them 4 times,sample space=HH,HT,TH,TT
for (a)P(hh)=1/4=0.25
for (b)P(ht)=2/4=0.50
hence these will remain same for 172 tosses also.

Answer 7

it's a statistics problem

a) = 172
b)=172

Answer 8

Why dont you try it yourself ?

answer to both is a chnace of 50% for either of them.

i aint the time for it

Answer 9

50/50 chances are

Answer 10

What is TWI?