Question

When a golfer plays any hole, he will take 3, 4, 5, 6, or 7 strokes with
probabilities of 1/10, 1/5, 2/5, 1/5,
and 1/10 respectively. He never takes more than 7 strokes. Find the probability of the following events:
a) scoring 4 on each of the first three holes
b) scoring 3, 4 and 5 (in that order) on the first three holes
c) scoring a total of 28 for the first four holes
d) scoring a total of 10 for the first three holes
e) scoring a total of 20 for the first three holes.​

Answer 1

(a) The probability of scoring 4 on each of the first three holes is 0.008.

(b) The probability of scoring 3, 4 and 5 (in that order) on the first three holes is 0.008.

(c) The probability of scoring a total of 28 for the first four holes is 0.0001.

(d) The probability of scoring a total of 10 for the first three holes is 0.006.

(e) The probability of scoring a total of 20 for the first three holes is 0.006.

Step-by-step explanation:

We are given with the following probability distribution of a golfer below;

   

         Strokes (X)                              Probability : P(X)  

               3                                                 $\frac{1}{10}$

               4                                                  $\frac{1}{5}$

               5                                                  $\frac{2}{5}$

               6                                                  $\frac{1}{5}$

               7                                                  $\frac{1}{10}$    

            Total                                              1      

(a) The probability of scoring 4 on each of the first three holes is given by;

     =  P(X = 3) $\times$ P(X = 3) $\times$ P(X = 3)

     =  $\frac{1}{5} \times \frac{1}{5} \times\frac{1}{5}$  

     =  $\frac{1}{125}$  = 0.008

(b) The probability of scoring 3, 4 and 5 (in that order) on the first three holes is given by;

     =  P(X = 3) $\times$ P(X = 4) $\times$ P(X = 5)

     =  $\frac{1}{10} \times \frac{1}{5} \times\frac{2}{5}$  

     =  $\frac{1}{125}$  = 0.008

(c) The probability of scoring a total of 28 for the first four holes is given by;

28 can be achieved in the first four holes only in one way that in each hole 7 has been scored, i.e;

    =  P(X = 7) $\times$ P(X = 7) $\times$ P(X = 7) $\times$ P(X = 7)

    =  $(\frac{1}{10})^{4}$

    =  $\frac{1}{10000}$  = 0.0001

(d) The probability of scoring a total of 10 for the first three holes is given by;

A Score of 10 in the first three holes can be achieved in three different ways;

• Score of 3 in the first hole, 3 in the second hole and 4 in the third hole.

• Score of 3 in the first hole, 4 in the second hole and 3 in the third hole.

• Score of 4 in the first hole, 3 in the second hole and 3 in the third hole.

   

So, the required probability will be ;

  =  $(\frac{1}{10} \times \frac{1}{10} \times\frac{1}{5})+(\frac{1}{10} \times \frac{1}{5} \times\frac{1}{10})+(\frac{1}{5} \times \frac{1}{10} \times\frac{1}{10})$

  =  $3 \times (\frac{1}{10} \times \frac{1}{10} \times\frac{1}{5})$

  =  $\frac{3}{500}$  =  0.006

(e) The probability of scoring a total of 20 for the first three holes is given by;

A Score of 20 in the first three holes can be achieved in three different ways;

• Score of 6 in the first hole, 7 in the second hole and 7 in the third hole.

• Score of 7 in the first hole, 6 in the second hole and 7 in the third hole.

• Score of 7 in the first hole, 7 in the second hole and 6 in the third hole.

   

So, the required probability will be ;

  =  $(\frac{1}{5} \times \frac{1}{10} \times\frac{1}{10})+(\frac{1}{10} \times \frac{1}{5} \times\frac{1}{10})+(\frac{1}{10} \times \frac{1}{10} \times\frac{1}{5})$

  =  $3 \times (\frac{1}{10} \times \frac{1}{10} \times\frac{1}{5})$

  =  $\frac{3}{500}$  =  0.006