probability 2?

Question

. If three persons are randomly selected , whats the probability that two were born on Monday and the other on Tuesday?

2. If Jack studies for tmr's test the prob. of getting a pass is 4/5 , if he doesnt study for it , the prob of getting a pass is 1/10. Given the prob. for Jack to study the test is 7/8 , find the prob. that he will fail the test .

更新:

3. Bag A contains black and white chess pieces.If one chess piece is drawn at random from the bag,the prob.of drawinga white chess piece and a black one are p and q respectively. (THE ANSWERS FOR P AND Q ARE 2/3 AND 1/3 RESPECTIVELY.)

(a)Bag B contains 6 white chess pieces and 8 black chess pieces.One chess piece is drawn at random from bag A and is put to Bag B,the one chess piece is drawn at random from Bag B.Find the prob.that the chess piece drawn from bag B is white. (THIS IS CHALLENGING!)

更新 2:

FOR NO.1,ANS=3/343

FOR NO.2,ANS=23/80

FOR NO.3,ANS=4/9

CHECK THE ANSWERS FIRST!AND EXPLAIN;)

SO MANY THANKS:D

Answer 1

最佳解答

1.

First, select 2 persons which were born on Monday [ C(3,2) (1/7)² ]

Second, select 1 person which was born on Tuseday [ C(1,1) (1/7) ]

The probability

= C(3,2) (1/7)² C(1,1) (1/7)

= 3/343

2.

Condition 1：Jack studies for tmr's test and fail the test

The probability = (7/8)(1 - 4/5)

Condition 2：Jack doesn't studies for tmr's test and fail the test

The probability = (1 - 7/8)(1 - 1/10)

∴ The probability

= (7/8)(1 - 4/5) + (1 - 7/8)(1 - 1/10)

= 23/80

3.

Bag A contains black and white chess pieces.If one chess piece is drawn at random from the bag,the prob.of drawing a white chess piece and a black one are 2/3 and 1/3 respectively.

Bag B contains 6 white chess pieces and 8 black chess pieces.One chess piece is drawn at random from bag A and is put to Bag B,the one chess piece is drawn at random from Bag B.Find the prob.that the chess piece drawn from bag B is white.

Condition 1：

1. Drawn a white chess piece from bag A [ (2/3) ]

2. Drawn a white chess piece from bag B [ (6+1)/(6+1+8) = (7/15) ]

Condition 2：

1. Drawn a black chess piece from bag A [ (1/3) ]

2. Drawn a white chess piece from bag B [ (6)/(6+8+1) = (6/15) ]

The probability

= (2/3)(7/15) + (1/3)(6/15)

= 4/9