Answer:

  • 7^4
  • dfrac{27}{343}

Step-by-step explanation:

  • we are given 7 alphabets out of which we need to choose 4 alphabets in order to make a code of 4 letters.

    Hence the size of the sample space is: 7^4

     since we have to fill in 4 places with the help of these 7 letters.

  so in first place we can select one out of any 7 letters, similarly for the second place as it is not mentioned that repetition is not allowed and the same can be done for third and fourth place.

Hence size of sample space is: 7times7times7times7=7^4

  • Now the probability of an event is given by:

 Probability=(Number of favourable outcomes)/(Total number of outcomes)

Hence here the total number of outcomes=7^4

Total number of favourable outcomes=outcomes that the first three letters are vowels=3times3times3times7 (since in the first place we have three choices for a vowel{A,I,0} similarly for second and third place and in the fourth place any one of the 7 letters could come)

Hence the probability(P) that the first three letters of the four-letter code are vowels is given by:

P=dfrac{3times3times3times7}{7times7times7times7}=dfrac{3times3times3}{7times7times7}=dfrac{27}{343}

Hence, the Probability is: dfrac{27}{343}