Question

The numbers 1, 2, 3, and 4 are written on slips of paper, and 2 slips are drawn at random one at a time without replacement. Find the given probabilities. a. The sum of the numbers is 7. b. The sum of the number is 3 or less. c. The first number is 2 or the sum is 4. a. The probability that the sum of the numbers is 7 is (Type an integer or a simplified fraction.)

Answer 1

Total cases for sum of these number is 12

A) P(sum of number is 7) = 1/6= 0.167

B) P( Sum of number is 3 Or less) = 1/6= 0.167

C) P(first number is 2 Or sum is 4) =

= 0.25+0.167=0.417

Solution file is attached go through it

Salt flymbers 1, 2, 3 and 4 are written paper and a slips are djawn at senden on glips of one at a time without replacement. find probe of sum of numbers is 7. Required favourable Cases Sum is 7 Prob. Total Cases. Coses. Total Cases = (1, 2) (23) Sumis 3 (1,2), (2,1) Cases Sum is u (13) (3,1) Sum is 5 (1,4) (4,1) (2,3) (3,2) sum is 6 (4,2) (2,4) Sum is 7 (4,3) (3,4) Total Ceses = 2 of 2 of 4+2+2 = 12 favourable Cases = Q =(4,3) (3,4) (Sum is 7) Bobo (sum is 7) = 12 E 0.167 6. 67 Bob. (Sum of numbe is 3 or less) = fertelosable Case Total Case Probe Sum is 3 or less = (1, 2) (211) 2 Coses

Boob. (Sum is 3 o less) = a I 0.167 6 Bobo (first number is 2 es sum is 4) = P(first Number is 2) + P(symis 4) - P(first 2 and sum is 4) P (first number is 2) ferrable lost No. 2 3 - I. 0.25 Total Cases 12 4. P( Sum is 4) = favourable Sum is 4 Q Ia 0.167 6 la Total Cages P(Ist Number 2 and Sum 4) = Bobo fest number is for -0.25 +0.167 - 0.417

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