A box contains 5 tickets numbered 1,2,3,4, and 5. - id5123455118845, drelisech6467

Ответ:

jeanniebyrd54

18.01.2022

(a) The probability that the numbers on the two tickets differ by two or more if the draws are made with replacement is 0.48.

(b) The probability that the numbers on the two tickets differ by two or more if the draws are made without replacement is 0.60.

Step-by-step explanation:

The tickets are drawn such that the difference between the two numbers is at least 2.

That is, 1st number - 2nd number ≥ 2.

The sample space such that this condition is satisfied is:

S = {(1, 3), (1, 4), (1, 5), (2, 4), (2, 5) and (3, 5)} = 6 possible pairs.

But there is also case where the numbers can be drawn in reverse order, i.e we can draw (3, 1) instead of (1, 3).

This makes the total number of possible pairs as, 6 × 2 = 12 pairs.

(a) With Replacement

In case the tickets are selected with replacement, then the probability of selecting the 1st ticket is same as for the 2nd ticket is:

$P(Difference\geq 2)=P(1st\ ticket)\times P(2nd\ ticket)\times No.\ of\ Possible\ pairs$

$=\frac{1}{5}\times \frac{1}{5}\times12\\ =0.48$

Thus, the probability that the numbers on the two tickets differ by two or more if the draws are made with replacement is 0.48.

(b) Without Replacement:

In case the tickets are selected without replacement, then the probability of selecting the 1st ticket is same as for the 2nd ticket is:

$P(Difference\geq 2)=P(1st\ ticket)\times P(2nd\ ticket)\times No.\ of\ Possible\ pairs$

$=\frac{1}{5}\times \frac{1}{4}\times12\\ =0.60$

Thus, the probability that the numbers on the two tickets differ by two or more if the draws are made without replacement is 0.60.

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