Question

A bag contains 12 two rupee coins, 7 one rupee coins and 4 half a rupee coins. If three
coins are selected at random then find the probability that (i) the sum of the three
coins is maximum (ii) the sum of the three coins is minimum (iii) each coin is of
different value.
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Answer 1

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Solution :-

Total number of coins in the bag

⠀⠀⠀⠀12 two rupee + 7 one

⠀⠀⠀= rupee coins + 4 half rupee coins

⠀⠀⠀= 23 coins

number of coins selected = 3

Experiment : selecting three coins from the bag

total number of possible choices

⠀⠀ number of ways in which

= three coins can be drawn from ⠀⠀⠀

⠀the total 23

⠀⠀⠀⠀⠀⠀⠀⇒n = ²³C₅

⠀⠀⠀⠀⠀⠀⠀⠀⠀ = $\small\rm{\dfrac{23 \times 22 \times 21 \times 20 \times 19}{ 5 \times 4 \times 3 \times 2 \times 1}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀ = 23 x 11 x 7 x 19

⠀⠀⠀⠀⠀⠀⠀⠀⠀ = 33,649

Let,

• A be the event of the sum of the three coins being maximum
• B be the event of the sum of the three coins be minimum
• C be the event of the three coin being different values

For Event A

The Sum of the 3 coins is maximum when the 3 coins chosen 2 rupee coin

Event A ⇒ Event of the selecting three 2 rupee coin

⠀⠀⠀⠀⠀⠀⠀⠀⠀2₹ ⠀ 1₹⠀⠀⠀ half ₹ ⠀⠀total

Available ⠀⠀12⠀⠀7 ⠀⠀⠀⠀⠀4⠀⠀⠀⠀23

To Choose ⠀⠀3⠀⠀0⠀⠀⠀⠀⠀0⠀⠀⠀⠀⠀3

choice⠀⠀⠀⠀¹²C₃⠀⁷C₀⠀⠀⠀⠀⁴C₀⠀⠀⠀²³C₃

No.of Favorable choices :-

⠀= the number of ways in which the ⠀⠀⠀

⠀3 two Rupees coin can be selected ⠀

⠀from the total 12 Two rupee coins

⠀⠀⠀⠀⇒ mₐ = ¹²C₃

⠀⠀⠀⠀⠀⠀⠀⠀= $\small\rm{\dfrac{12 \times 11 \times 10 }{ 3 \times 2 \times 1}}$

⠀⠀⠀⠀⠀⠀⠀⠀= 220

probability of the sum of the three coins being maximum

⇒ probability of occurrence of event A

= number of favourable choices of the ⠀

event : total number of possible choices ⠀

for the experiment

p(A) ⇒mₐ/n

⠀⠀⠀= 220/33,649

⠀⠀⠀= 20/3,059

▪ Odds

Number of unfavourable choices

⠀⠀⠀⠀= total number of possible choices - ⠀⠀⠀⠀⠀⠀number of favourable choices

>> In favour

odds in favour of the sum of the three coins being maximum

⇒odds in favour of event A

⠀⠀⠀⠀⠀= number of unfavourable choices : ⠀⠀⠀⠀⠀⠀number of favourable

⠀⠀⠀⠀⠀= mₐ : mₐᶜ

⠀⠀⠀⠀⠀= 220 : 33,429

⠀⠀⠀⠀⠀= 20 : 3,039

Against

odds in against of the sum of the three coins being maximum

⇒odds in favour of event A

⠀⠀⠀⠀⠀= number of unfavourable choices : ⠀⠀⠀⠀⠀⠀number of favourable

⠀⠀⠀⠀⠀= mₐᶜ : mₐ

⠀⠀⠀⠀⠀= 33,429 : 220

⠀⠀⠀⠀⠀= 3,039 : 20

For Event B

The Sum of the 3 coins is minimum when the 3 coins chosen Half rupee coin

Event B ⇒ Event of the selecting three half rupee coin

⠀⠀⠀⠀⠀⠀⠀⠀⠀2₹ ⠀ 1₹⠀⠀⠀ half ₹ ⠀⠀total

Available ⠀⠀12⠀⠀ 7 ⠀⠀⠀⠀4⠀⠀⠀⠀⠀23

To Choose ⠀⠀0⠀⠀0⠀⠀⠀⠀⠀3⠀⠀⠀⠀⠀3

choice⠀⠀⠀⠀¹²C₀⠀⠀⁷C₀⠀⠀⠀⁴C₃⠀⠀⠀²³C₃

No.of Favorable choices :-

= the number of ways in which the ⠀⠀⠀⠀

3 half Rupees coin can be selected ⠀⠀⠀

from the total 12 half rupee coins

⠀⠀⠀⠀⇒ mₐ = ⁴C₃

⠀⠀⠀⠀⠀⠀⠀⠀= ⁴C ₄ - ₃

⠀⠀⠀⠀⠀⠀⠀⠀= ⁴C₁

⠀⠀⠀⠀⠀⠀⠀⠀= 4

probability of the sum of the three coins being minimum

⇒ probability of occurrence of event B

= number of favourable choices of the

event : total number of possible choices ⠀

for the experiment

$\rm{p(B) ⇒m_b/n}$

⠀⠀⠀= 4/33,649

For Event C

⠀⠀⠀⠀⠀⠀⠀⠀⠀2₹ ⠀ 1₹⠀⠀⠀ half ₹ ⠀⠀total

Available ⠀⠀12⠀⠀7 ⠀⠀⠀⠀4⠀⠀⠀⠀⠀23

To Choose ⠀⠀1⠀⠀ 1⠀⠀⠀⠀⠀1⠀⠀⠀⠀⠀3

choice⠀⠀⠀⠀¹²C₁⠀⠀⁷C₁⠀⠀⠀⁴C₁⠀⠀⠀²³C₃

No.of Favorable Choice :-

⠀= The no of ways in which the a two rupee coin , a one rupee coin and half rupee coin can be selected from total 23 coins

mₐ ⇒ ²C₁ x ⁷C₁ x ⁴C₁⠀

⠀⠀ = 12 x 7 x 4

⠀⠀ = 336

probability of the three coins being different value

⇒ probability of occurrence of event C

⠀= number of favourable choices of the

event : total number of possible

choices for the experiment

$\rm{p(C) ⇒m_c/n}$

⠀⠀⠀= 336/33,649

$\large\sf{\red{@\:Aꪀցꫀl}}$

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Answer 2

Answer:

The bag contains 12 two rupee coins, 7 one rupee coins and 4 half a rupee coins.

Total coins in the bag =12+7+4=23

Selecting three coins from the bag =

23

C

3

=1771

Sum of three coins will be maximum when all the three coins selected are of rupee two.

Therefore,

Selecting three coins of rupee two =

12

C

3

=220

Probability of selecting three coins with maximum sum =

1771

220

=

161

20

Sum of three coins will be minimum when all the three coins selected are of half a rupee coin.

Therefore,

Selecting three coins of half a rupee =

4

C

3

=4

Probability of selecting three coins with minimum sum =

1771

4

Selecting three different coins =12

1

C

×

7

C

1

×

4

C

1

=12×7×4=336

Probability of selecting three different coins from the bag =

1771

336

=

253

48

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