Question

When a golfer plays any hole, he will take 3, 4, 5, 6, or 7 strokes withprobabilities 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 holesb) scoring 3, 4 and 5 (in that order) on the first three holesc) scoring a total of 28 for the first four holesd) scoring a total of 10 for the first three holese) 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)  P(X = 3)  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)  P(X = 4)  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)  P(X = 7)  P(X = 7)  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