#### Q. No. 6 :

Three identical
skipping ropes are dropped in a heap. The six free ends of the ropes
are picked up

by three people—each
person picks up one free end with his right hand and another free end
with his
left. Then they all step away from each other.

(a) What is the
probability that every person finds he is holding a single skipping
rope?

(b) What is the
probability that the three people are joined in a single closed
chain?

#### Answer :

**(a) What is the probability that every person finds he is holding a single skipping rope?**

There are 3 ropes.

Each rope has two
ends.

Therefore, there are
a total of 6 ends.

Let us call the rope
ends as follows :

Rope Ends

1 1 – 2

2 3 – 4

3 5 - 6

Let us assume that
the first person catches rope 1- end 1.

Now he has got 5
more ends left.

The probability of
getting end 2 is therefore P(A) = 1/5

Let us now assume
that the second person catches rope 2- end 3.

Now he has got 3
more ends left.

The probability of
getting end 4 is therefore P(B) = 1/3

Let us assume that
the third person catches rope 3- end 5.

Now he has got only
1 more end left.

The probability of
getting end 6 is therefore P(C) = 1/1= 1

Therefore the total
probability is P = P(A)*P(B)*P(C)= 1/5*1/3*1=1/15

The python code for
the same is given below :

```
```

```
```

`tie1=0`

`attempt1=0`

`a=0`

`for rop1 in range(1,7):`

` if rop1==1:`

` continue`

` for rop2 in range(1,6):`

` if rop2==1:`

` tie1+=1`

` attempt1+=1`

`a+=float(tie1)/attempt1`

tie2=0

`attempt2=0`

`b=0`

`for rop3 in range(1,5):`

` if rop3==1:`

` continue`

` for rop4 in range(1,4):`

` if rop4==1:`

` tie2+=1`

` attempt2+=1`

`b+=float(tie2)/attempt2`

tie3=0

`attempt3=0`

`c=0`

`for rop5 in range(1,3):`

` if rop5==1:`

` continue`

` for rop6 in range(1,2):`

` if rop6==1:`

` tie3+=1`

` attempt3+=1`

`c+=float(tie3)/attempt3`

print('The probability = ',float(a*b*c))

####
```
``````
```

Answer :

```
```

**(b) What is the probability that the three people are joined in a single closed chain?**

Let us assume that
the first person catches rope 1- end 1.

Now he has got 4 out of 5 ends left as he should not choose end 2.

Therefore the
probability is P(A) = 4/5

Let us now assume
that the second person catches rope 2- end 3.

Now he has got 2 out of 3 ends left as he should not choose end 4.

Therefore the
probability is P(B) = 2/3

Now the third person
has got only 1 more option left.

Therefore the
probability is P(C) = 1/1= 1

Therefore the total
probability is P = P(A)*P(B)*P(C)= 4/5*2/3*1=8/15

The python code is given below :

The python code is given below :

```
s1=0
```

tries1=0

a=0

for first1 in range(1,7):

if first1==1:

continue

for first2 in range(2,7):

if first2==2:

s1+=1

tries1+=1

a+=1-float(s1)/tries1

s2=0

tries2=0

b=0

for second1 in range(3,7):

if second1==3:

continue

for second2 in range(4,7):

if second2==4:

s2+=1

tries2+=1

b+=1-float(s2)/tries2

s3=0

tries3=0

c=0

for third1 in range(5,7):

if third1==1:

continue

for third2 in range(6,7):

if third2==2:

s3+=1

tries3+=1

c+=1-s3/tries3

print('The probability = ',float(a*b*c))
` `

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