Question

a dice is thrown twice find the probability of getting first same number second different numbers third consecutive numbers fourth alternative numbers

Answer 1

$\Huge{\underline{\underline{\mathfrak{Question \colon}}}}$

A dice is thrown twice find the probability of getting :

a) Same Numbers

b) Different Numbers

c) Consecutive Numbers

d) Alternate Numbers

$\Huge{\underline{\underline{\mathfrak{Answers \colon }}}}$

A dice is thrown twice.

Total no.of outcomes would be : 36

Outcomes would be :

$\boxed{\begin{minipage}{7 cm} \large{\tt{11 \ 12 \ 13 \ 14 \ 15 \ 16}} \\ \large{\tt{21 \ 22 \ 23 \ 24 \ 25 \ 26}} \\ \large{\tt{31 \ 32 \ 33 \ 34 \ 35 \ 36}} \\ \large{\tt{41 \ 42 \ 43 \ 44 \ 45 \ 46}} \\ \large{\tt{51 \ 52 \ 53 \ 54 \ 55 \ 56}} \\ \large{\tt{61 \ 62 \ 63 \ 64 \ 65 \ 66}} \end{minipage}}$

Using the Formula,

$\huge{ \boxed{ \sf{p(e) = \frac{favourable \: outcomes}{total \: no.of \: outcomes} }}}$

a) Let A be the given emperical event

• A = Same Numbers i.e, 11,22,33,.......,66

Now,

P(A) = 6/36

→ P(A) = 1/6

b) Let B be the given event

• B = Different Numbers i.e., 16,12 etc

$\huge{\boxed{ \sf{p( \bar{e}) = 1 - p(e)}}}$

Implies,

P(B) = 1 - P(A)

→ P(B) = 1 - 1/6

→ P = 5/6

c) Let C be the given emperical event

• C = Consecutive Numbers like 12,23,.......,56

Now,

P(C) = 5/36

d) Let D be the given emperical event

• D = Alternate Numbers i.e., 13,24,35,46,64,53,42 and 31 » 10

Now,

P(D) = 10/36

→ P(D) = 5/18

Note

• Probability of an event should always be expressed in simple fractions

Answer 2

$\large{\underline{\underline{\mathfrak{\blue{\sf{Correct\:question-}}}}}}$

A dice is thrown twice. Find the probability of getting-

1. Same number
2. Different numbers
3. Consecutive numbers
4. Alternative numbers

$\large{\underline{\underline{\mathfrak{\blue{\sf{Answer-}}}}}}$

★ Probability of getting a same number = $\sf{\dfrac{1}{6}}$

★ Probability of getting different numbers = $\sf{\dfrac{5}{6}}$

★ Probability of getting consecutive numbers = $\sf{\dfrac{5}{36}}$

★ Probability of getting alternate numbers = $\sf{\dfrac{5}{18}}$

$\large{\underline{\underline{\mathfrak{\blue{\sf{Explanation-}}}}}}$

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Outcomes :

(1,1) ⠀⠀(1,2) ⠀⠀(1,3) ⠀⠀(1,4) ⠀⠀(1,5) ⠀⠀(1,6)

(2,1) ⠀⠀(2,2) ⠀⠀(2,3) ⠀⠀(2,4) ⠀⠀(2,5) ⠀⠀(2,6)

(3,1) ⠀⠀(3,2) ⠀⠀(3,3) ⠀⠀(3,4) ⠀⠀(3,5) ⠀⠀(3,6)

(4,1) ⠀⠀(4,2) ⠀⠀(4,3) ⠀⠀(4,4) ⠀⠀(4,5) ⠀⠀(4,6)⠀⠀

(5,1) ⠀⠀(5,2) ⠀⠀(5,3) ⠀⠀(5,4) ⠀⠀(5,5) ⠀⠀(5,6)

(6,1) ⠀⠀(6,2) ⠀⠀(6,3) ⠀⠀(6,4) ⠀⠀(6,5) ⠀⠀(6,6)

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$\large{\underline{\boxed{\mathfrak{\sf{\pink{Total\:Number\:of\:Outcomes=36}}}}}}$

General formula for finding probability of an event :

$\large{\underline{\boxed{\mathfrak{\sf{\green{P(e)=\dfrac{Favourable\:Outcomes}{Total\:outcomes}}}}}}}$

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★A dice is thrown twice, probability of getting the same number : $\bold{\sf{\dfrac{1}{6}}}$

Here, favourable outcomes = (1,1); (2,2); (3,3); (4,4); (5,5); (6,6)

Favourable outcomes = 6

Probability of getting a same number :

$\sf{\dfrac{Favourable\:outcomes}{Total\:Outcomes}}$

$\implies$ Probability of getting a same number = $\sf{\cancel{\dfrac{6}{36}}}$

$\implies$ Probability of getting a same number = $\red{\sf{\dfrac{1}{6}}}$

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★ A dice is thrown twice, probability of getting different numbers : $\bold{\sf{\dfrac{5}{6}}}$

Here, favourable outcomes = 36 - 6 = 30

Probability of getting different numbers :

$\sf{\dfrac{Favourable\:outcomes}{Total\:Outcomes}}$

Probability of getting different numbers = $\sf{\cancel{\dfrac{30}{36}}}$

Probability of getting different numbers = $\red{\sf{\dfrac{5}{6}}}$

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★ A dice is thrown twice, probability of getting consecutive numbers : $\bold{\sf{\dfrac{5}{36}}}$

Here, favourable outcomes = (1,2); (2,3); (3,4); (4,5); (5,6)

Favourable outcomes = 5

Probability of getting consecutive numbers :

$\sf{\dfrac{Favourable\:outcomes}{Total\:Outcomes}}$

Probability of getting consecutive numbers = $\red{\sf{\dfrac{5}{36}}}$

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★ A dice is thrown twice, probability of getting alternate numbers : $\bold{\sf{\dfrac{5}{18}}}$

Here, Alternate numbers = (1,3); (2,4); (3,5); .....

Favourable outcomes = 10

Probability of getting alternate numbers :

$\sf{\dfrac{Favourable\:outcomes}{Total\:Outcomes}}$

Probability of getting alternate numbers = $\sf{\cancel{\dfrac{10}{36}}}$

Probability of getting alternate numbers = $\red{\sf{\dfrac{5}{18}}}$

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#answerwithquality

#BAL