Question

Five coins whose faces are marked 2 3 are tossed the chance of obtaining a total of 12 is

Answer 1

Answer:

The answer is 5/16.

Step-by-step explanation:

As we know that  

Favorable case→5!/3!2!

⇒5⋅4⋅3⋅2/3⋅2⋅2=10

Total case=2^5=32

P=10/32=5/16.

Hence the probability of obtaining a total of 12 is 5/16.

Answer 2

The chance of obtaining a total of 12 is $\dfrac{5}{16}$

Step-by-step explanation:

Given : Five coins whose faces are marked 2 3 are tossed.

Outcomes on each coin =2

Total outcomes on tossing 5 such coins = $2^5=32$

To make sum as 12 , we need combination of three 2's and two 3's.

The number of combinations of having three 2's (other will be 3's) = $^5C_3$

$=\dfrac{5!}{(5-3)!3!}=\dfrac{5\times4}{2}=10$

i.e. The number of favorable outcomes = 10

Now , The chance of obtaining a total of 12 =$\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$

$=\dfrac{10}{32}=\dfrac{5}{16}$

∴ The chance of obtaining a total of 12 is $\dfrac{5}{16}$ .

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