Recently, I came across a probability related worksheet here expected to be solved by students of African Institute for Mathematical Sciences. The were expected to solve the problems using Random function of Python.

Here I would like to show you an attempt of solving those problems by a beginner.

#### Q. No. 1:

1. Suppose you have a fair six-sided die. Now suppose that you throw the die three times. What is the

probability that you will get:

(a) a 1 on the first throw, a 2 on the second throw, and a 3 on the third throw?

(b) a 1, a 2, and a 3 in any order?

(c) three sixes?

probability that you will get:

(a) a 1 on the first throw, a 2 on the second throw, and a 3 on the third throw?

(b) a 1, a 2, and a 3 in any order?

(c) three sixes?

#### Answer :

(a)Generally, the probability of obtaining 1 on first through is 1/6 as there are 6 possibilities.

Similarly, the possibility of obtaining 2 on second through and 3 on third through is also 1/3 each.

If P(A) =P(B) = P(C) = 1/6

Then, the total probability P = P(A) * P(B) * P(C) = 1/6 ³

The python code is shown below :

`success = 0`

tries = 0

for first in range (0,6) :

for second in range (0,6) :

for third in range (0,6):

if first == 1 and second == 2 and third == 3:

success+=1

tries+=1

print (float(success)/tries)

Answer = 0.004629629629629629 i.e. 1/6^3

(b)

a 1, a 2, and a 3 in
any order?

Answer :

The number of
possibilities of getting 1, 2 and 3 in any order are :

123

132

213

231

312

321

Hence, there are 6
possibilities.

But the total
possibilities are 111 to 666. i. e. 6^3

Therefore, the total
probability is 6/6^3 = 1/6^2

The python code for
this problem is shown below :

```
success = 0
```

tries = 0

for first in range(0,6):

for second in range(0,6):

for third in range(0,6):

if first == 1 :

if second == 2 :

if third == 3:

success+=1

elif second == 3:

if third == 2 :

success+=1

elif first == 2 :

if second == 1 :

if third == 3:

success+=1

elif second == 3:

if third == 1 :

success+=1

elif first == 3 :

if second == 1 :

if third == 2:

success+=1

elif second == 2:

if third == 1 :

success+=1

tries+=1

print (float(success)/tries)

```
#Ans : 0.027777777777777776 i.e. 6/6^3 = 1/6^2
```

(c)

The probability of
getting a six on first or second or third die is 1/6

The total
probability is the sum of all these probabilities.

If P(A) =P(B) = P(C)
= 1/6

Then, the total
probability P = P(A) * P(B) * P(C) = 1/6 ³

The python code is
shown below :

`success = 0`

tries = 0

for first in range(1,7):

for second in range(1,7):

for third in range(1,7):

if first == 6 and second == 6 :

if third == 6 :

success+=1

tries+=1

print ('Success=',success)

print ('tries=',tries)

print ('Probability=',float(success)/tries)

#Ans = 0.004629629629629629 i.e. 1/6^3

```
```

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