A student has a 25% chances to be accepted by the University A, 40% by University B?

Question

What is the probabability to pe accepted by University A and rejected by B?

Answer 1

1/4 to be accepted by uni A and 3/5 to be rejected by B

so you multiply 1/4 with 3/5 which equals to 4/20

so 20%

Answer 2

mango's?

Answer 3

Hi,

Accept..Not

25%.....75%.UnivA

-----------------------

25x40..|.75x40...| Accept Univ B

------------------------|

25x60..|.75x60...| Not Univ B

-------------------------

Accept..Not

25%.....75%.UnivA

-----------------------

_.1___|_.3_____| Accept Univ B

------------------------|

_.15__|_.45____| Not Univ B

-------------------------

25% accepted at univ A x 60% not accepted at univ B = 15% of the time <==ANSWER

I hope that helps!! :-)

Answer 4

42.5 percent

Answer 5

Basically he has a 25% chance of being accepted at A.  And he has a 60% chance of being rejected at B.

P(A) = 0.25

P(B) = 0.40

P(not B) = 1 - P(B) = 0.60

The probability of both events happening is the product:

P(A and not B) = P(A) * P(not B)

= 0.25 * 0.60

= 0.15

Answer:

15%