there are two urn containing (4 red and 5 black balls) and (5 red and 6 black baltherels ). one ball is drawn at random from the first and transfer to the second and thenone ball is drawn from second and transfer to first. after the transfer one ball is drawn from the first urn , what is the chance that it will be the red ball ?
Answers
U1 = 5 R & 6 W. U2= 3 R & 7 W.
Transfer 2 balls from U1 to U2 to get
U31 +2R Probability (5/11)×(4/10)=(2/11).
5R&7W.
U32 +2W Probability (6/11)×(5/10)=(3/11). 3R&9W.
U33 +WR Or +RW Probability (6/11)×(5/10) + (5/11)×(6/10)=(6/11). 4R&8W.
Above three cover all four outcomes for 2 balls from U1 to U2 without replacement.
Finally Requested Probability of getting 2 or both red balls from possible U2 as modified to U31 or U32 or U33 using conditional Probability concept = ( Probability (U31) × P(RR | U31) ) + ( (P(U32) × P(RR | U32) ) + ( (P(U33) × P(RR | U33))
= ((2/11)×(5/12)×(4/11)) + ((3/11)×(3/12)×(2/11)) + ((6/11)×(4/12)×(3/11))
= (10/363) + (3/242) + (6/121)
=(1/726) × (20+9+36)
= 65/726
= 0.0895316804 in decimals. Answer
Answer:
Step-by-step explanation:
Total number of balls in 1st urn = 4 + 5 = 9
Total number of balls in 2nd urn = 5 + 6 = 11
One ball is picked from the 1st urn and dropped into the 2nd urn. Similarly, one ball is taken from the 2nd and dropped into the 1st. This makes the number of balls equal all the time.
Hence,
Total number of red balls = 4 + 5 = 9
Total number of black balls = 5 + 6 = 11
Probability of picking up the red ball from the first urn = 9/9 = 1