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4. Roll two dice together and then add the two scores.
(a) find the total possible scores
(b) Can you say the results are equally likely outcomes?
(c) Use your table to find the probability of getting a total
score of
(i) 10 (ii) 1 (iii) 16 (iv) 12 (v) less than 6
​

Sagot :

Answer:

See the below work.

Step-by-step explanation:

To find the outcomes of adding the scores when rolling two dice together, first we make the table of outcomes - refer to the 1st attached picture.

(a)

the total possible scores (outcome) = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(b)

The results are not equally likely outcomes since the total numbers of each outcome are unequal:

[tex]\begin{array}{c|c}\cline{1-2}outcome&total\ (frequency)\\\cline{1-2}2&1\\3&2\\4&3\\5&4\\6&5\\7&6\\8&5\\9&4\\10&3\\11&2\\12&1\\\cline{1-2}\Sigma frequencey&36\\\cline{1-2}\end{array}[/tex]

(c)

We can find the probability by using the probability formula:

[tex]\boxed{P(A)=\frac{n(A)}{n(S)} }[/tex]

where:

  • [tex]P(A)=\text{probability of event A}[/tex]
  • [tex]n(A)=\text{number of outcome of event A}[/tex]
  • [tex]n(S) = \text{total number of all outcomes}[/tex]

(i)

Given:

  • [tex]n(x=10)=3[/tex]
  • [tex]n(S)=36[/tex]

Then:

[tex]\begin{aligned} P(x=10)&=\frac{n(x=10)}{n(S)}\\\\&=\frac{3}{36} \\\\&=\bf\frac{1}{12} \end{aligned}[/tex]

(ii)

Given:

  • [tex]n(x=1)=1[/tex]
  • [tex]n(S)=36[/tex]

Then:

[tex]\begin{aligned} P(x=1)&=\frac{n(x=1)}{n(S)}\\\\&=\bf\frac{1}{36}\end{aligned}[/tex]

(iii)

Given:

  • [tex]n(x=16)=0[/tex]
  • [tex]n(S)=36[/tex]

Then:

[tex]\begin{aligned} P(x=16)&=\frac{n(x=16)}{n(S)}\\\\&=\frac{0}{36} \\\\&=\bf0 \end{aligned}[/tex]

(iv)

Given:

  • [tex]n(x=12)=1[/tex]
  • [tex]n(S)=36[/tex]

Then:

[tex]\begin{aligned} P(x=12)&=\frac{n(x=12)}{n(S)}\\\\&=\bf\frac{1}{36} \end{aligned}[/tex]

(v)

Given:

  • [tex]n(x < 6)=1+2+3+4=10[/tex]
  • [tex]n(S)=36[/tex]

Then:

[tex]\begin{aligned} P(x < 6)&=\frac{n(x < 6)}{n(S)}\\\\&=\frac{10}{36} \\\\&=\bf\frac{5}{18} \end{aligned}[/tex]

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