Question

Suppose two dice are rolled. Find the probabilities of the following events a) the maximum of the two numbers rolled is less than or equal to 2; b) the maximum of the two numbers rolled is less than or equal to 3; c) the maximum of the two numbers rolled is exactly equal to 3 d) Repeat b) and c) for r instead of 3, for each z from 1 to 6. e) Denote by P(x) the probability that the maximum number is exactly r. What should P(1) P(2) answers to d). P(3) P(4) P(5) P(6) equal

Answer 1

Answer:

The maximum of the two numbers rolled is less than or equal to $2$ is $\frac{1}{9}$

The maximum of the two numbers rolled is less than or equal to 3 is $\frac{1}{4}$

The maximum of the two numbers rolled is exactly equal to 3 is $\frac{5}{36}$

Step-by-step explanation:

Step 1 of 5

a) the maximum of the two numbers rolled is less than or equal to $2$

outcome                                $X$

$(1,1)$                                  $1$

$(1,2)(2,1)(2,2)$                       $2$

The probability $= 4/36$

$= 1/9$

Step 2 of 5

b) the maximum of the two numbers rolled is less than or equal to $3$

outcome                          $X$

$(1,1)$                           $1$

$(1,2)(2,1)(2,2)$                   $2$

$(1,3)(3,1)(2,3)(3,2)(3,3)$   $3$

Total outcome $=9$

The probability $=9/36$

$= 1/4$

Step 3 of 4

c) the maximum of the two numbers rolled is exactly equal to 3

outcome                           $X$

$(1,3)(3,1)(2,3)(3,2)(3,3)$   $3$

The probability $= 5/36$

Step 4 of 5

d) Repeat b) and c) for r instead of 3, for each z from 1 to 6.

for b)

let two numbers rolled is less than or equal to X

$X$        $P(X)$

$\leq 1$      $1/36$

$\leq 2$      $4/36$

$\leq 3$      $9/36$

$\leq 4$     $16/36$

$\leq 5$     $25/36$

$\leq 6$     $36/36$

for c)

$X$ be highest of two values  

outcome                                                                          $X$      $P(X)$

$(1,1)$                                                                           $1$       $1/36$

$(1,2)(2,1)(2,2)$                                                            $2$       $3/36$

$(1,3)(3,1)(2,3)(3,2)(3,3)$                                            $3$       $5/36$

$(1,4)(4,1)(2,4)(4,2)(3,4)(4,3)(4,4)$                           $4$          $7/36$

$(1,5)(5,1)(2,5)(5,2)(3,5)(5,3)(4,5)(5,4)(5,5)$            $5$       $9/36$

$(1,6)(6,1)(2,6)(6,2)(3,6)(6,3)(4,6)(6,4)(5,6)(6,5)(6,6)$   $6$        $11/36$

Step 5 of 5

e) Denote by P(x) the probability that the maximum number is exactly r

$X$  $P(X)$

$1$    $1/36$

$2$    $3/36$

$3$     $5/36$

$4$         $7/36$

$5$     $9/36$

$6$       $11/36$

$P(I) +P(2) +P(3) +P(4) +P(5) +P(6) = 36/36$

$= 1$

Learn more about probability, refer :

https://brainly.com/question/13604758