Question

PLEASE RESPOND.

Rita is making a beaded bracelet. She has a collection of 160 blue beads, 80 gray beads, and 240 pink beads. What is the estimated probability that Rita will need to pick at least five beads before she picks a gray bead from her collection?


(A 0.05
(B 0.10
(C 0.45
(D 0.55

Answer 1

Given that Rita is making a beaded bracelet. She has a collection of 160 blue beads, 80 gray beads, and 240 pink beads. We are to calculate the probability that Rita will need to pick atleast 5 beads before she picks a grey bead from her collection.

Prob for drawing atleast 5 beads before she picks a grey bead from her collection

= 1-Prob for drawing atleast one grey beed in the first 5 draws.

(Because these two are complementary events)

no of grey beeds drawn in first 5 trials is

Bin

Prob for drawing atleast one grey beed in the first 5 draws.

=1-Prob of no grey

Hence required prob=P(X=0 in first 5 draws)

=

6th beeds onwards can be grey also.

Nearest answer is c)0.45

Explanation:

Answer 2

Explanation:

Given that Rita is making a beaded bracelet. She has a collection of 160 blue beads, 80 gray beads, and 240 pink beads. We are to calculate the probability that Rita will need to pick atleast 5 beads before she picks a grey bead from her collection.

Prob for drawing atleast 5 beads before she picks a grey bead from her collection

= 1-Prob for drawing atleast one grey beed in the first 5 draws.

(Because these two are complementary events)

no of grey beeds drawn in first 5 trials is

Bin
`(5,(80)/(80+160+240) \\=(5,(1)/(6) )`

Prob for drawing atleast one grey beed in the first 5 draws.

=1-Prob of no grey

Hence required prob=P(X=0 in first 5 draws)

=
`(1-(1)/(6) )^5\\=0.4018`

6th beeds onwards can be grey also.

Nearest answer is c)0.45