Question

STAT_14_3

Ronit has a box with beads. The beads are opaque or transparent and available in several colors.
The probability that a random bead will be red is 0.3. The probability that a bead will be transparent is 0.6.
Of the red beads - the probability of a random bead being transparent is 0.5

A. Remove 8 beads from the box at random and upon return (sampling with replacement). What is the probability that exactly two of them will be red?
B. Take beads out of the box accidentally and on return (sampling with replacement) until you first remove a transparent bead.
1. What is the probability that more than 4 beads will be removed?
2. The first two beads taken out were not transparent. What is the probability of getting 7 beads out of the box?
C. Remove 10 beads from the box at random and upon return (sampling with replacement). What is the probability that exactly three of them will be red and transparent, two opaque and red and 5 transparent and red?

Answer 1

SOLUTION:

Given that the probability that a random bead will be red is 0.3. The probability that a bead will be transparent is 0.6.

a)

With replacement random sampling with probablity success = 0.3

where total number of trials is 8

so,if x is number of red beads then

b) 1)

Let x be the number of trails before setting the 1st transpernent bead

here

so, p[x>=4] { i.e; x+1>4}

=1-0.9744

=0.0256

2)

Geometric distribution has the memoriless property

so p[x=6 | x>=2] {note: 7 bounds out of the box means 6 bounds not transparent that's why x=63}

=p[x=4]

=(0.6)(0.4)⁴

=0.1536

c)

(a) 3 = Red and Transparent

(b) 2= Red and Opaque

(c) 5 = Transparent and opaque

8 = Red and Transparent

2 = Red and Opaque

(as 'a' and 'c' are same we just add them to make 8).

If we draw 10 balls, then they can be

1) Red and Transparent

2) Red and Opaque

3) Not Red

this multinominal distribution when Prob of distibution when prob of case (1) is =0.3*0.5 = 0.15(P₁)

prob of case (2) is =0.3*0.5 = 0.15(P2)

prob of case (3) is = 0.7(P₃)

So, by multinominal distribution, the prob is

.

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