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Answer 1

$\underline{\underline{\huge{\textbf{Solution:}}}}$

Given,
A bag contains slips bearing numbers from 1- 89.

Experiment - A Slip is drawn at random from the bag

Let $E_{1}$ be the Event that drawn slip has the number such that $\textit{sum of the digits is 11}$

Let $E_{2}$ be the Event that the drawn slip has a $\textit{number greater than 63}$

$\underline{\Large{\textsf{Step-1:}}}$
Find total number of outcomes when a slip is drawn at random from bag

Total No. of Outcomes = No. of ways of drawing a slip from bag containing 89 slips

$\implies \textbf{Total\: No.\: of\: outcomes = 89}$

$\underline{\Large{\textsf{Step-2:}}}$
Find No. of Favorable outcomes for Occurrence of Event $E_{1}$$[n(E_{1})]$

Favorable outcomes = No. of Numbers from 1- 89 having sum of digits as 11

Now,
Find the Numbers whose sum of digits is 11

The Sum of digits is 11, implies it is a $\textit{2-digit Number from 11 to 89}$ without having $\textsf{zero}$ as one of its digit

$\textsf{Sum of digits for a two digit No. = Ten's digit + Unit's digit}$

Let Ten's digit be x

$\implies x+ Unit's digit = 11$

$\implies Unit's digit = 11-x$

$\underline{\Large{Form\: the\: Number\: using\: the\: digits}}$

$\textsf{A Two-digit No. is equal to sum of 10 times the Ten's digit and Unit's digit}$

$2-digit\: No. = 10(Ten's digit)+1(Unit's digit)$

By Substituting

$\implies 2-digit\: No. = 10(x)+(11-x)$

$\implies \underline{\textsf{2-digit No. = 9x+11}}$

Substitute x= 2 to x=8 to get the Numbers whose sum of digits is 11

$\textbf{ E_{1} = \{29, 38, 47, 56, 65, 74, 83\}}$

$\implies \textbf{No.\: of\: Favorable\: outcomes\: n( E_{1} ) = 7}$

$\underline{\Large{\textsf{Step-3:}}}$
Find No. of Favorable outcomes for Occurrence of Event $E_{2}$$[n(E_{2})]$

No. of Favorable outcomes = No. of Numbers from 64 to 89

No. of No.s from 64 to 89 = 89-64+1 = 25+1= 26

$\textbf{ E_{2} = \{64, 65, 66, 67, .... , 88, 89\}}$

$\implies \textbf{No.\: of\: Favorable\: outcomes\: n( E_{2} ) = 26}$

$\underline{\Large{\textsf{Step-4:}}}$
Find the Probability of occurrence of Events
$E_{1}$ & $E_{2}$

We have,
$\bigstar \mathbf{Probability\: of\: Occurrence\: of\: Event = \dfrac{No.\: of\: favorable\: outcomes\: for\: Event}{Total\: No.\: of\: outcomes\: in\: Experiment}}$

$P(E_{1}) = \dfrac{n(E_{1})}{n(S)}$

Substituting values

$\implies P(E_{1}) = \dfrac{7}{89}$

$P(E_{2}) = \dfrac{n(E_{2})}{n(S)}$

Substituting values

$\implies P(E_{2}) = \dfrac{26}{89}$

$\therefore$
$\bigstar\: \textsf{Probability that drawn slip has a number whose sum of it's digit is 11 = \underline{\large{\textbf{ \dfrac{7}{89} }}}}$

$\: \: \: \, \textsf{Probability that drawn slip has a number greater than 63 = \underline{\large{\textbf{ \dfrac{26}{89} }}}}$

Answer 2

Solution:

Given,

A bag contains slips bearing numbers from 1- 89.

Experiment - A Slip is drawn at random from the bag

$Let E_{1}E </p><p>1$

be the Event that drawn slip has the number such that \textit{sum of the digits is 11}sum of the digits is 11

$Let E_{2}E </p><p>2$

be the Event that the drawn slip has a $\textit{number greater than 63}$

$\underline{\Large{\textsf{Step-1:}}}$

Find total number of outcomes when a slip is drawn at random from bag

Total No. of Outcomes = No. of ways of drawing a slip from bag containing 89 slips

$\implies \textbf{Total\: No.\: of\: outcomes = 89}$

$\underline{\Large{\textsf{Step-2:}}}$

Find No. of Favorable outcomes for Occurrence of

$[n(E_{1})][n(E </p><p>1</p><p> </p><p> )]$

Favorable outcomes = No. of Numbers from 1- 89 having sum of digits as 11

Now,

Find the Numbers whose sum of digits is 11

$The Sum of digits is 11, implies it is a \textit{2-digit Number from 11 to 89}2-digit Number from 11 to 89 without having \textsf{zero}zero as one of its digit</p><p>$

$\textsf{Sum of digits for a two digit No. = Ten's digit + Unit's digit}Sum of digits for a two digit No. = Ten’s digit + Unit’s digit</p><p>$

Let Ten's digit be x

$\implies x+ Unit's digit = 11⟹x+Unit </p><p>′</p><p> sdigit=11$

$\implies Unit's digit = 11-x⟹Unit </p><p>′</p><p> sdigit=11−x$

$\underline{\Large{Form\: the\: Number\: using\: the\: digits}}$

$\textsf{A Two-digit No. is equal to sum of 10 times the Ten's digit and Unit's digit}$

2-digit\: No. = 10(Ten's digit)+1(Unit's digit)[/tex](Ten

′

sdigit)+1(Unit

′

sdigit)

By Substituting

$\implies 2-digit\: No. = 10(x)+(11-x)}$

$\implies \underline{\textsf{2-digit No. = 9x+11}}$

Substitute x= 2 to x=8 to get the Numbers whose sum of digits is 11

$\textbf{ E_{1} = \{29, 38, 47, 56, 65, 74, 83\}}E </p><p>1$

= {29, 38, 47, 56, 65, 74, 83}

$\implies \textbf{No.\: of\: Favorable\: outcomes\: n( E_{1} ) = 7}$

$\underline{\Large{\textsf{Step-3:}}}$

Step-3:

Find No. of Favorable outcomes for Occurrence of Event E_${2}E$

No. of Favorable outcomes = No. of Numbers from 64 to 89

No. of No.s from 64 to 89 = 89-64+1 = 25+1= 26

$\textbf{ E_{2} = \{64, 65, 66, 67, .... , 88, 89\}}$

$\implies \textbf{No.\: of\: Favorable\: outcomes\: n( E_{2} ) = 26}$

$\underline{\Large{\textsf{Step-4:}}}$

Find the Probability of occurrence of Events

$E_{1}E$

We have,

$\bigstar \mathbf{Probability\: of\: Occurrence\: of\: Event = \dfrac{No.\: of\: favorable\: outcomes\: for\: Event}{Total\: No.\: of\: outcomes\: in\: Experiment}}$

$P(E_{1}) = \dfrac{n(E_{1})}{n(S)}$

Substituting values

$\implies P(E_{1}) = \dfrac{7}{89}$

$P(E_{2}) = \dfrac{n(E_{2})}{n(S)}$

Substituting values

$\implies P(E_{2}) = \dfrac{26}{89}}$

$\therefore∴$

$\bigstar\: \textsf{Probability that drawn slip has a number whose sum of it's digit is 11 = [tex]\underline{\large{\textbf{ \dfrac{7}{89} }}}}$

$\: \: \: \, \textsf{Probability that drawn slip has a number greater than 63 = \underline{\large{\textbf{ \dfrac{26}{89} }}}}$