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Cambridge Maths Academy
6. Probability - Jumping goldfish (GCSE) 본문
6. Probability - Jumping goldfish (GCSE)
Cambridge Maths Academy 2020. 12. 29. 00:23
수학 모음 (Maths collection) 전체보기
There are goldfish in two bowls P and Q.
- During one minute the probability of one goldfish jumping from bowl P to Q is 25.
- During one minute the porbability of one goldfish jumping from bown Q to P is 13.
Calculate the probability that:
(a) the number of goldfish in each bowl at the end of 1 minute is equal to the number of goldfish in each bowl at the start of the minute;
(b) the number of goldfish in each bowl at the end of 2 minutes is equal to the number of goldfish in each bowl at the start of this two minute period.
(Higher GCSE Maths 4-9 by Michael White, E8.3 Mixed probability questions Q10.)
Solution. (a) We have P(P→Q)=25P(P↛Q)=35 and P(P←Q)=13P(P↚Q)=23 Note: We do not consider the cases where more than one goldfish jump from one bowl to the other, we simply don't. However, if one's interested in multiple goldfish jumping, it can be developed into a Poisson distribution.
In order for each bowl to have the same number as the start after one minute, there are two possibilities. (1) One goldfish jumps from P to Q and one goldfish jumps from Q to P. That is, P1=P(P→Q)×P(P←Q)=25×13=215 (2) No goldfish jump, i.e. P2=P(P↛Q)×P(P↚Q)=35×23=25 We find P=P1+P2=215+25=815
(b) In order for each bowl to have the same number as the start after two minutes, the followings are possible.
- (1) P→Q in the 1st minute, and P←Q in the 2nd minute.
- (2) P←Q in the 1st minute, and P→Q in the 2nd minute.
- (3) P⇄Q in the 1st minute, and nothing happens in the 2nd minute.
- (4) Nothing happens in the 1st minute, and P⇄Q in the 2nd minute.
- (5) P⇄Q in the 1st minute, and P⇄Q in the 2nd minute.
- (6) Nothing happens in the 1st minute, and nothing happens in the 2nd minute.
We find the respective probabilities:
- P→Q in the 1st minute, and P←Q in the 2nd minute: Note that 'P→Q in the 1st minute' means that P→Q and P↚Q in the 1st minute; and similarly 'P←Q in the 2nd minute' means that P↛Q and P←Q in the 2nd minute, i.e. P1=P(P→QandP↚Q)1st×P(P↛QandP←Q)2nd=(25×23)×(35×13)=415×15=475
- P←Q in the 1st minute, and P→Q in the 2nd minute: P2=P(P←QandP↛Q)1st×P(P→QandP↚Q)2nd=(13×35)×(25×23)=15×415=475
- P⇄Q in the 1st minute, and nothing happens in the 2nd minute: P3=P(P→QandP←Q)1st×P(P↛QandP↚Q)2nd=(25×13)×(35×23)=215×25=475
- Nothing happens in the 1st minute, and P⇄Q in the 2nd minute: P4=P(P↛QandP↚Q)1st×P(P→QandP←Q)2nd=(35×23)×(25×13)=25×215=475
- P⇄Q in the 1st minute, and P⇄Q in the 2nd minute: P5=P(P→QandP←Q)1st×P(P→QandP←Q)2nd=(25×13)×(25×13)=215×215=4225
- Nothing happens in the 1st minute, and nothing happens in the 2nd minute: P6=P(P↛QandP↚Q)1st×P(P↛QandP↚Q)2nd=(35×23)×(35×23)=25×25=425
We find P=6∑r=1Pr=475+475+475+475+4225+425=88225
Aside. For a one-minute period, there are four possible scenarios:
- P→Q
- P←Q
- P⇄Q
- Nothing happens.
For a two-minute period, there are 42=16 possibilities as follows.

- P→Q in the 1st minute, and P→Q in the 2nd minute. (P decreases by 2 and Q increases by 2.) P1=P(P→QandP↚Q)1st×P(P→QandP↚Q)2nd=(25×23)×(25×23)=415×415=16225
- P→Q in the 1st minute, and P←Q in the 2nd minute. (P,Q have the same number as the start.) P2=P(P→QandP↚Q)1st×P(P←QandP↛Q)2nd=(25×23)×(13×35)=415×15=475
- P→Q in the 1st minute, and P⇄Q in the 2nd minute. (P decreases by 1 and Q increases by 1.) P3=P(P→QandP↚Q)1st×P(P→QandP←Q)2nd=(25×23)×(25×13)=415×215=8225
- P→Q in the 1st minute, and nothing happens in the 2nd minute. (P decreases by 1 and Q increases by 1.) P4=P(P→QandP↚Q)1st×P(P↛QandP↚Q)2nd=(25×23)×(35×23)=415×25=875
- P←Q in the 1st minute, and P→Q in the 2nd minute. (P,Q have the same number as the start.) P5=P(P←QandP↛Q)1st×P(P→QandP↚Q)2nd=(13×35)×(25×23)=15×415=475
- P←Q in the 1st minute, and P←Q in the 2nd minute. (P increases by 2 and Q decreases by 2.) P6=P(P←QandP↛Q)1st×P(P←QandP↛Q)2nd=(13×35)×(13×35)=15×15=125
- P←Q in the 1st minute, and P⇄Q in the 2nd minute. (P increases by 1 and Q decreases by 1.) P7=P(P←QandP↛Q)1st×P(P→QandP←Q)2nd=(13×35)×(25×13)=15×215=275
- P←Q in the 1st minute, and nothing happens in the 2nd minute. (P increases by 1 and Q decreases by 1.) P8=P(P←QandP↛Q)1st×P(P↛QandP↚Q)2nd=(13×35)×(35×23)=15×25=225
- P⇄Q in the 1st minute, and P→Q in the 2nd minute. (P decreases by 1 and Q increases by 1.) P9=P(P→QandP←Q)1st×P(P→QandP↚Q)2nd=(25×13)×(25×23)=215×415=8225
- P⇄Q in the 1st minute, and P←Q in the 2nd minute. (P increases by 1 and Q decreases by 1.) P10=P(P→QandP←Q)1st×P(P←QandP↛Q)2nd=(25×13)×(13×35)=215×15=275
- P⇄Q in the 1st minute, and P⇄Q in the 2nd minute. (P,Q have the same number as the start.) P11=P(P→QandP↚Q)1st×P(P→QandP←Q)2nd=(25×13)×(25×13)=215×215=4225
- P⇄Q in the 1st minute, and nothing happens in the 2nd minute. (P,Q have the same number as the start.) P12=P(P→QandP←Q)1st×P(P↛QandP↚Q)2nd=(25×13)×(35×23)=215×25=475
- Nothing happens in the 1st minute, and P→Q in the 2nd minute. (P decreases by 1 and Q increases by 1.) P13=P(P↛QandP↚Q)1st×P(P→QandP↚Q)2nd=(35×23)×(25×23)=25×415=875
- Nothing happens in the 1st minute, and P←Q in the 2nd minute. (P increases by 1 and Q decreases by 1.) P14=P(P↛QandP↚Q)1st×P(P←QandP↛Q)2nd=(35×23)×(13×35)=25×15=225
- Nothing happens in the 1st minute, and P⇄Q in the 2nd minute. (P,Q have the same number as the start.) P15=P(P↛QandP↚Q)1st×P(P→QandP←Q)2nd=(35×23)×(25×13)=25×215=475
- Nothing happens in the 1st minute, and nothing happens in the 2nd minute. (P,Q have the same number as the start.) P16=P(P↛QandP↚Q)1st×P(P↛QandP↚Q)2nd=(35×23)×(35×23)=25×25=425

For changes in the number of goldfish in the bowls, there are five cases.
- P decreases by 2 and Q increases by 2: P(2↓,2↑)=P1=16225
- P decreases by 1 and Q increases by 2: P(1↓,1↑)=P3+P4+P9+P13=8225+875+8225+875=64225
- P,Q have the same number as the start: P(0↓,0↑)=P2+P5+P11+P12+P15+P16=475+475+4225+475+475+425=88225
- P increases by 1 and Q decreases by 2: P(1↑,1↓)=P7+P8+P10+P14=275+225+275+225=1675
- P increases by 2 and Q decreases by 2: P(2↑,2↓)=P6=125
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