Question

2 cards numbered 1, 2, 3,......11,12 are kept in a box and mixed thoroughly. If one card is drawn at random, find the probability of getting a card with: (a) a prime number (b) a factor of 12 (c) a number divisible by 3 (d) a multiple of 2

Answer 1

Hey mate!

Total possible outcomes=12

1. P(getting a prime number)=5/12=(2,3,5,7,11)
2. P(getting a factor of 12)=6/12=1/2(1,2,3,4,6,12)
3. P(getting a number divisible by 3)=4/12=1/3(3,6,9,12)

P(getting a multiple of 2)=6/12=1/2(2,4,6,8,10,12) thank uu

Answer 2

TOTAL CONDITION = 12

A) A prime number

FAVOURABLE CONDITIONS= 5{2,3,5,7,11}

$\boxed{PRIBABLITY=\frac{FAVOURABLE \: CONDITION }{TOTAL \: CONDITION }}$

$PROBABLITY=\dfrac{5}{12}$

B) A factor of 12

FAVOURABLE CONDITIONS= 6 {1,2,3,4,6,12}

$\boxed{PRIBABLITY=\frac{FAVOURABLE \: CONDITION }{TOTAL \: CONDITION }}$

$PROBABLITY=\dfrac{6}{12}$

$PROBABLITY=\dfrac{1}{2}$

C) A number divisible by 3

FAVOURABLE CONDITIONS= 4{3,6,9,12}

$\boxed{PRIBABLITY=\frac{FAVOURABLE \: CONDITION }{TOTAL \: CONDITION }}$

$PROBABLITY=\dfrac{4}{12}$

$PROBABLITY=\dfrac{1}{3}$

C) A number multiple 2

FAVOURABLE CONDITIONS= 6{1,4,6,8,10,12}

$\boxed{PRIBABLITY=\frac{FAVOURABLE \: CONDITION }{TOTAL \: CONDITION }}$

$PROBABLITY=\dfrac{6}{12}$

$PROBABLITY=\dfrac{1}{2}$