Question

STAT 14_1:

In the jug there are balls of different colors.
The probability of accidentally taking out a red-ball from the jug is P
When (0<P<1)

a. Remove from the jug, incidentally and with return, six balls.
The probability that exactly one ball came out red is four times greater than the probability that no red balls came out.
i. Find the p.
ii. What is the probability that only the first, third, and fifth balls will be red?

b. Now they decided to take balls out of the jug, incidentally and with return, until the first red ball came out.
i. What is the probability of getting exactly five balls out of the jug?
ii. What does the number of non-red balls expect from the jug?

c. Yoav decided to take balls out of the pitcher, incidentally and with return, until he removed 4 red balls (not necessarily in succession)
What is the probability that he will get exactly 9 balls?

Answer 1

Solution

Since the draw is with return, the probability remains the same for each draw.

P(red ball) = P => P(non-red ball) = 1 - P

Part (a)

Sub-part (i)

Exactly one red ball => other 5 are non-red

One red ball can be drawn in any one of 6 draws in 6 ways. Thus,

P(Exactly one red ball) = 6P(1 - P)⁵

P(None of 6 balls is red) = (1 - P)⁶

By the given stipulation,

6P(1 - P)⁵ = 5(1 - P)⁶ [Note: four times greater => 5 times]

=> P = 5/11 Answer 1

Sub-part (ii)

P(only the first, third, and fifth balls will be red)

= P(1 - P) x P(1 - P) x P(1 - P)

= {(5/11)(6/11)}³

= 0.0152 Answer 2

Part (b)

Sub-part (i)

Given condition => first four were non-red and then the fifth was red.

So, the required probability

= (1 - P)⁴ x P

= (5/11) x (6/11)⁴

= 0.0402 Answer 3

Sub-part (ii)

Let X = number of non-red balls drawn before the first red ball is drawn. Then,

X = 0 with probability P

X = 1 with probability (1 - P)P

X = 2 with probability (1 - P)²P

X = 3 with probability (1 - P)³P

And so on, mathematically this could go to infinity.

Thus, expected number of non-red balls

= (0 x P) + {1 x (1 - P)P} + {2 x (1 - P)²P} + {3 x (1 - P)³P} + ........... to infinity

= (1 - P)/P [using the sum formula for infinite Arithmetico-geometric series]

= 6/5 = 1.2 Answer 4

Part (c)

The given condition => the last ball, i.e., the 9^th ball must be red and out of the preceding 8 balls, 3 must be red and 5 must be non-red in any order.

So, the required probability

= (⁸C₃)(5/11)³(6/11)⁵(5/11)

= 56(5/11)⁴(6/11)⁵

= 0.1154 Answer 5

DONE