You play two games against the same opponent. The probability you win the first game is 0.7. If you win the first game, the probability you also win the second is 0.5. If you lose the first game, the probability that you win the second is 0.3.(a) Are the two games independent?(b) What's the probability you lose both games?

Given Information:

Probability of wining 1st game = p₁ = 0.7

Probability of wining 2nd game given 1st game won = p₂|p₁ = 0.5

Probability of wining 2nd game given 1st game lost = p₂|q₁ = 0.3

Required Information:

(a) Are the two games independent = ?

(b) Probability of losing both games = ?

(a) Are the two games independent = No

(b) Probability of losing both games = 0.21

Step-by-step explanation:

(a) Independent Events:

Two events are said to be independent when the success of one event is not affected by the success or failure of another event.

In this case, the probability of 2nd game depends on the success or failure of the 1st game, therefore, the two games are not independent.

(b) Probability of losing both games

The probability of losing the both games is the product of the probabilities of losing each game.

Probability of losing 1st game = 1 - Probability of wining 1st game

Probability of losing 1st game = 1 - 0.7 = 0.30

Probability of losing 2nd game = 1 - Probability of wining 2nd game given 1st game lost

Probability of losing 2nd game = 1 - 0.3 = 0.70

Please note that since we are finding the probability of losing both games that's why we used the condition of 1st game lost

Probability of losing both games = Probability of losing 1st game*Probability of losing 2nd game

Probability of losing both games = 0.30*0.70

Probability of losing both games = 0.21

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47