, 16.05.2020kenn14

I have two bags of counters. The first bag contains 2 RED counters and 1 BLUE counters. The second bag contains 1 RED counter, 1 BLUE counter and 2 YELLOW counters. I take a counter at random from both bags.1. What is the probability that the two counters will be the same color?2. What is the probability that exactly one of the counters will be red?​

Step-by-step explanation:

1) First bag {r,r,b} Second bag {r,b,y,y}

S={(r,r),(r,b),(r,y),(r,y),(r,r),(r,b),(r,y),(r,y),(b,r),(b,b),(b,y),

(b,y)}

= 12

Total number of counters with the same color,

=(r,r),(r,r),(b,b)

= 3

Pr(2 counters of the same color)

= 3/12 = ¼ = 25%

2) Total number of counters with only one red,

=(r,b),(r,y),(r,y),(r,b),(r,y),(r,y),(b,r)

= 7

Pr(counter with only one red)

= 7/12

ok I understand this questions

1. If red counter = 3/7

If blue counter = 2/7

2. 3/7

3/12, i used a tree diagram to work it out

ok I understand this questions

1. If red counter = 3/7

If blue counter = 2/7

2. 3/7

Step-by-step explanation:

1) First bag {r,r,b} Second bag {r,b,y,y}

S={(r,r),(r,b),(r,y),(r,y),(r,r),(r,b),(r,y),(r,y),(b,r),(b,b),(b,y),

(b,y)}

= 12

Total number of counters with the same color,

=(r,r),(r,r),(b,b)

= 3

Pr(2 counters of the same color)

= 3/12 = ¼ = 25%

2) Total number of counters with only one red,

=(r,b),(r,y),(r,y),(r,b),(r,y),(r,y),(b,r)

= 7

Pr(counter with only one red)

= 7/12

blue mixedyellow to be red

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