Answer:
The probability of selecting randomly 6 packets such that the first 4 contain cocaine and 2 not is 0.3522.
Step-by-step explanation:
Probabilities of selecting 4 packets with the ilegal substance:
=
Combinassions possible= 1365
Probabilities of selecting 2 packets with white powder:
Combinations possible= 10
Probabilities of selecting 6 packets from the totality of them:
Combinations possible= 38760
The probability of picking 4 with the substance and 2 with only white powder is:
= 0.3522
I hope this answer helps you.
A=3B
2 for $0.25
1.10/ 6= .20 (rounding)
1.00/ 5= .20
0.85/ 4= .21
0.25/ 2= .13 (rounding)