pls nairalanders I definately know u people can solve this problem, help me out pls I was given this assignment since last week and unfortunately I tried all my possible best and trick to solve it but I cant so I want U to give it a try. moderator pls help and put in a front page.
"in any trial of an experiment a rat may turn left or right, if it turns left it receive food, if it turns right it receive an electric shock on the first trial the rat may equally likely turn left or right, if the rat receive food on any trial the probability of left turn to the next trial is 1/5, if the rat receive an electric shock on any trial the probability of left turn to the next trial is 1/3, the rat turns three times.
1) What is the probability of receiving an electric shock and two food.
2) what is the probability of recieving three electric shock.
3) what is the probability of recieving three food.
Thanx 4 ur participating.

waiting 4 ur response

If the rat turn to eat food,the probality left to turn is 1/5 and if it turn to other way{shock}the probability left we be 1/3. To calculate the probability for the two trial left is 1/5-1=4/5 note when it turn to eat food. When it turn to shock 1/3-1=2/3. No1= 4/5*2/3 =8/15. No2= 2/3*3=2. No3= 4/5*3 =12/5.

Since the rat has an equal chance of turning left or right on the first trial, the probability of every first turn is = 1/2.
The probability of turning to food after receiving food in the previous turn is 1/5, therefore the probability of turning to an electric shock after receiving food in the previous turn is 1-1/5 =4/5.
The probability of turning to food after recieving an electric shock in d previous turn is 1/3, therefore the probability of turning to an electric shock again after recieving a shock in the previous turn is 1-1/3=2/3.
Thus the solutions are
1. Probability of getting two food and one shock = P(FSF) + P(SFF) + P(FFS). Where F represents food and S represents shock.
Thus, probability = (1/2*4/5*1/3) + (1/2*1/3*1/5) + (1/2*1/5*4/5) = 2/15 + 1/30 + 2/25 = 37/150
2. The probability of recieving three electric shock = P(S&S&S) = (1/2*2/3*2/3)= 2/9.
3. The probability of recieving three food = P(F&F&F)= (1/2*1/5*1/5)=1/50.

oh thanks very much 4 ur responses may God blessing ang mercies be upon u all of that response. LONG LIVE NAIRALAND.

since prob. for d first trials are equal,then then prob for left and right for first trial is 1/2 resp.
if it recieves food on any trial then prob. for nxt turn to the left is 1/5.then prob. to d right is 1-1/5=4/5,
if it recieve shock on any trial then prob. of nxt turn to the right is1/3,then to the right is 1-1/3=2/3,
1)f,f,s ,f,s,f ,s,f,f
(1/2*1/5*4/5)+(1/2*4/5*1/3)+(1/2*1/3*1/5)=37/150
2) s,s,s
1/2*2/3*2/3=2/9
3) f,f,f
1/2*1/5*1/5=1/50
@4tyse7en u re absolutly correct
@hayourbam no probability question is equal to 1 or greater than 1 but its usually lesser than 1

@martins thanx very much for taking ur time solve the problem 4me I am absolutely glad that I had enough from u guys