Q. 2.24

Question

If two dice are rolled, what is the probability that the sum of the upturned faces equals $i ?$ Find it for $i = 2, 3, . . ., 11, 12 .$

Step-by-Step Solution

Verified

Answer

$P a r t (a) \frac{1}{3} [2 ex] P a r t (b) \frac{1}{2}$

1Step 1 Given Information.

Consider the given information:

If two dice are rolled and the the value of i is 2, 3, ..., 11, 12.

2Step 2 Explanation.

Considered evens:

$G_{1}$ -the first of the two balls are painted gold.

$G_{2}$ -the second of the two balls is painted gold.

$P (G_{1}) = \frac{1}{2} \Rightarrow P (G_{1}^{C}) = \frac{1}{2} P (G_{2}) = \frac{1}{2} \Rightarrow P (G_{2}^{C}) = \frac{1}{2}$

The evens $G_{1}$ $&$ $G_{2}$ are independent.

3Step 3 Part (a) Explanation.

$P (G_{1} G_{2} ∣ (G_{1} \cup G_{2})) = \frac{P [G_{1} G_{2} (G_{1} \cup G_{2})]}{P (G_{1} \cup G_{2})}$

$G_{1} G_{2} (G_{1} \cup G_{2}) = G_{1} G_{2} G_{1} \cup G_{2} = {(G_{1}^{C} G_{2}^{c})}^{c}$

$P (G_{1} G_{2} | (G_{1} \cup G_{2})) = \frac{P (G_{1} G_{2})}{P [{(G_{1}^{C} G_{2}^{c})}^{c}]} = \frac{P (G_{1}) P (G_{2})}{1 - P (G_{1}^{c} G_{2}^{c})} I n d e p e n d e n c e = \frac{P (G_{1}) P (G_{2})}{1 - P (G_{1}^{c}) P (G_{2}^{c})} I n d e p e n d e n c e$

$= \frac{\frac{1}{2} \cdot \frac{1}{2}}{1 - \frac{1}{2} \cdot \frac{1}{2}} = \frac{1}{3}$

4Step 4 Explanation.

One specific ball is golden, what is the probability that the other one is golden?

Either $G_{1}$ occurs, and the wanted probability is, $P (G_{2} ∣ G_{1})$ or $G_{2}$ occurs, and the wanted probability is $P (G_{1} ∣ G_{2})$ .

From the definition of independence,

$P (G_{1} ∣ G_{2}) = P (G_{1}) = \frac{1}{2}$

$P (G_{2} ∣ G_{1}) = P (G_{2}) = \frac{1}{2}$

This is different because in $a)$ we know that one of the three events happened, and here there are only two possible outcomes - by the color of the second ball.

Q. 2.23

Q. 2.27

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