Question

- A bag contains 13 balls numbered from 1 to
13. An even number is considered a 'success'.
Two balls are drawn, with replacement, from
the bag. Find the probability of getting:
(a) Two successes (b) exactly one success
(c) at least one success (d) no success.​

Answer 1

Answer:

36/169

Step-by-step explanation:

Given that drawing an even number is success out of total 13 possible outcomes so

P(S)=

13

6

Probability of failure =P(F)=

13

7

Probability of getting two success out of two drawn =(

13

6

)

2

=

169

36

.

Answer 2

The probability of getting:

(a) Two successes  = 36/169

(b) exactly one success = 84/169

(c) at least one success  = 120/169

(d) no success = 49/169

Given:

A bag contains 13 balls numbered from 1 to 13.

An even number is considered a 'success'.

Two balls are drawn, with replacement, from the bag

To Find:

The probability of getting:

(a) Two successes

(b) exactly one success

(c) at least one success

(d) no success.​

Solution:

• Probability of an event  = n(E)/n(S)  
• n(E) = number of possible outcome of event
• n(S) = number of possible sample space outcome
• P(E) + P('not E') = 1

Step 1:

Total Numbers from 1 to 13 are 13

Even Numbers from 1 to 13 are  6   ( 2 , 4 , 6 , 8 , 10 , 12)

P ( getting Even number) = 6/13

P ( not getting Even Number) = 1 - 6/13 = 7/13

Step 2:

Calculate probability of Getting 2 success

= (6/13)(6/13)

= 36/169

Step 3:

Calculate probability of Getting exactly one success

1st Even number 2nd odd number or 1st Odd number 2nd Even Number

= (6/13)(7/13) + (7/13)(6/13)

= (42 + 42)/169

= 84/169

Step 4:

Calculate probability of Getting at least one success

= 1 - P ( no success)

= 1 - (7/13)(7/13)

= 1 - 49/169

= (169 - 49)/169

= 120/169

Step 5:

Calculate probability of Getting no success

both even number

 (7/13)(7/13)

=  49/169