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Multiplication and addition principle for count possibilities?/

  • Thread starter rclakmal
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multiplication and addition principle for count possibilities???????/

we toss a fair coin 5 times and recored the sequence of head and tails obtained.how may sequences are possible that start with head OR end with head.??????

so what i did was using addition theorm i divided this complex process in to 3 simple partitions like below

X1=start with head AND end with head
X2=start with head AND end with tail
X3=start with tail AND end with head

so my answer should be X1+X2+X3........

but im supposed to find X1,X2,and X3 by using the multiplication theorem but im little bit confused about how to use it

can anyone help me out.thanks!~!!!!!!!!!!
 

Answers and Replies

  • #2
jbunniii
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In each case, you have constrained exactly two of the five coin flips.

If you had constrained all five, then of course there would be only one sequence matching the constraints.

If you had constrained four out of five, then you have one degree of freedom (one specific coin toss would be allowed to be a head or a tail), so there would be two sequences satisfying the constraints.

Can you use a similar line of reasoning for the case where exactly two of the coin flips (the first and the last) are constrained?
 
  • #3
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yr thanks it helps me a lot i can get the answe using that ,,,,,,,,,,

but can u help me about multiplication thorem it says if a process separated to several stages where there are M1 choices in stage 1 ,M2 choices in stage 2...............Mr choces in stage r then i can get the total number of number of ways by

M1*M2*.....Mr

do u know about it and how to connect that with above problem ..............
 
  • #4
jbunniii
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yr thanks it helps me a lot i can get the answe using that ,,,,,,,,,,

but can u help me about multiplication thorem it says if a process separated to several stages where there are M1 choices in stage 1 ,M2 choices in stage 2...............Mr choces in stage r then i can get the total number of number of ways by

M1*M2*.....Mr

do u know about it and how to connect that with above problem ..............
Yes. First, let's consider tossing N coins in a row. How many possible distinct outcomes are there? You have N stages, and each stage has 2 possible outcomes, so the total number of possible sequences is just [tex]2^N[/tex].

Now, if you are asked to count the number of sequences where the first and the last coin toss are constrained, then if you think about that a bit, that's actually equivalent to simply tossing (N-2) normal coins, plus 2 coins that only have heads (or only have tails). So then M=2 for (N-2) of the coins, and M=1 for the other 2 coins. Then you have

[tex]2^{N-2} \cdot 1^2 = 2^{N-2}[/tex]

possible outcomes.

Another way to think about it is that it's equivalent to tossing a sequence of (N-2) coins, whereas the remaining two coins were already tossed by someone else prior to the experiment.
 
  • #5
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thanks a lot u saved my life .....:)..........by using that i got the answer as 24........

X1=8,X2=8,X3=8;;;;;;;;;;;;so
X=X1+X2+X3
X=24

thanks again ......!!!!!!!!!!!
 

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