Ok I gotta I’ll help you right now
<u>Step-by-step explanation:</u>
We have , ,
We know that
Substituting values of P & B ,
Now ,
⇒ ××2
⇒
⇒
⇒
⇒
Answer:
Its D.
Step-by-step explanation:
All you have to do is divide 3 1/2 by 1/2.
Since there is no repetition allowed, there are 10 possibilities for the 1st digit, 9 for the 2nd, 8 for the 3rd, and 7 for the 4th. This gives a total of (10)(9)(8)(7) = 5040 four-digit codes.
For all odd digits to be used, there are 5 possibilities for the 1st digit (1,3,5,7,9), 4 for the 2nd, 3 for the 3rd, 2 for the 4th. This gives a total of (5)(4)(3)(2) = 120 codes that only use odd digits.
Therefore there are 5040 - 120 = 4920 codes that do not consist of all odd digits. The probability is 4920/5040 = 41/42.
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