Question

bag contains slips on which English alphabet a,b,c... x,y,z are written. if one sleep is taken out, what is the probability that the slip contains
a) vowel
b) a constant
c) the letter m ​

Answer 1

Step-by-step explanation:

hope it helps you

5/26

21/26

1/26

Answer 2

• a) The probability that the slip contains a vowel = 5/26

• b) The probability that the slip contains a constant = 21/26

• c) The probability that the slip contains the letter m = 1/26

Given :

• A bag contains slips on which English alphabet a,b,c... x,y,z are written.

• One sleep is taken out

To find :

• a) The probability that the slip contains a vowel

• b) The probability that the slip contains a constant

• c) The probability that the slip contains the letter m

Solution :

Step 1 of 4 :

Write down total number of possible outcomes

Here it is given that the bag contains slips on which English alphabet a,b,c... x,y,z are written

Total number of English alphabets = 26

So total number of possible outcomes = 26

Step 2 of 4 :

Find the probability that the slip contains a vowel

One sleep is taken out

Let A be the event that the slip contains a vowel

Vowels are a , e , i , o , u

Total number of vowels = 5

So total number of possible outcomes for the event A is 5

∴ The probability that the slip contains a vowel

= P(A)

$\displaystyle \sf{ = \frac{Number \: of \: favourable \: cases \: to \: the \: event \: A }{Total \: number \: of \: possible \: outcomes }}$

$\displaystyle \sf{ = \frac{5}{26} }$

Step 3 of 4 :

Find the probability that the slip contains a constant

One sleep is taken out

Let B be the event that the slip contains a constant

Total number of consonants = 26 - 5 = 21

So total number of possible outcomes for the event B is 21

∴ The probability that the slip contains a constant

= P(B)

$\displaystyle \sf{ = \frac{Number \: of \: favourable \: cases \: to \: the \: event \: B }{Total \: number \: of \: possible \: outcomes }}$

$\displaystyle \sf{ = \frac{21}{26} }$

Step 4 of 4 :

Find the probability that the slip contains the letter m

One sleep is taken out

Let C be the event that the slip contains the letter m

Total number of "the letter m " = 1

So total number of possible outcomes for the event C is 1

∴ The probability that the slip contains a vowel

= P(C)

$\displaystyle \sf{ = \frac{Number \: of \: favourable \: cases \: to \: the \: event \: C}{Total \: number \: of \: possible \: outcomes }}$

$\displaystyle \sf{ = \frac{1}{26} }$

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