Question

A bag contains 4 five rupee coins, 3 two rupee coins and 2 one rupee coins. If 3 coins are drawn from the bag at random, what is the probability of the draw yielding maximum amount? Select one:

a. 5/46

b. 1/21

c. 3/43

d. 4/45

Answer 1

Answer:

c

Step-by-step explanation:

Answer 2

b. 1/21

The probability of the draw yielding maximum amount= $\dfrac{1}{21}$

Explanation:

Given : A bag contains 4 five rupee coins, 3 two rupee coins and 2 one rupee coins.

Total coins = 4+3+2=9

If 3 coins are drawn from the bag at random, then the , maximum amount would be the sum of 3 five rupee coins (as 5 rupee coin carries maximum value.)

The combination of r things taken out of n things = $^nC_r=\dfrac{n!}{r!(n-r)!}$

Using Combinations, the number of ways to draw 3 five rupee coins= $^4C_3=4$

[∵$^nC_{n-1}=n$]

Favorable outcomes = 4

The number of ways to draw any 3 coins = $^9C_3=\dfrac{9!}{3!6!}$

$=\dfrac{9\times8\times7\times6!}{6\times6!}=84$

Probability of the draw yielding maximum amount =$\dfrac{Favorable \ outcomes}{Total \ outcomes}$

$\dfrac{4}{84}=\dfrac{1}{21}$

So , the probability of the draw yielding maximum amount= $\dfrac{1}{21}$

Hence, the correct answer is b. $\dfrac{1}{21}$

# Learn more :

A bag contains 23 number of coins in the form of 1 rupee 2 rupee and 5 rupee coins the total sum of the coin is 43 .the ratio between 1 rupee and 2 rupee. coin is 3:2 then the number of 1 rupee coin is

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