Abag contains 12 counters. 7 of them are white. a counter is taken at random and not replaced. a second counter is taken out of the bag at random. calculate the probability that only one of the two counters is white

➷ First calculate how many counters are 'not white'

12 - 7 = 5

We have 5 counters that are 'not white'

The probability of taking a white counter the first time is 7/12

For the next pick, there would be 1 less of the total counters as it is not being replaced

The probability of then taking a 'not white' counter is 5/11

The other way you could still get one white counter is:

The first counter (non white) probability would be 5/12

The second counter (white) probability would be 7/11

Multiply these values:

7/12 x 5/11 = 35/132

5/12 x 7/11 = 35/132

Add these two values together:

35/132 + 35/132 = 70/132

Your answer is 70/132

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

well what you have to do is divide 84 by 6 so theanswer

step-by-step explanation:

84 divided by 6 = 14

and that lightbulb changing party must be freaking lit!

what is the information provided?

step-by-step explanation: