Question

6. A box contains 30 slips numbered from 1 to 30 and another box contains 50 slips numbered
from1 to 50. One slip is taken from each box.
a) In how many many different ways we can select a pair of slips, one from each box?
b) What is the probability of both being odd ?
c) What is the probability of getting an even number and an odd number ?
?
d) What is the probability of getting atleast an odd number​

Answer 1

Answer:

the answer of his questions is 440 cm square

Answer 2

a) No. of ways we can select from first box= 30C1

= 30 (30!/29!×1!= 30)

and from second box= 50C1= 50

Thus no. of ways we can select in pair

= 30×50= 1500

b)

Odd numbers in first box= 30/2= 15 odd numbers

odd numbers in second box=50/2=25

P(odd)= 15/30×25/50= 1/2×1/2= 1/4

c) Lets take it to the simplest solution, since there are only 2 groups i.e even or odd

total outcomes= 4 (even-even, even-odd,odd-even, odd-odd)

ways of getting even number and odd= 2

P(even&odd)= 2/4= 1/2

d) Take example from c again,

chances of getting at least an odd is (even-odd,odd-even and odd-odd)= 3

here at least one odd is present

so probability= 3/4