1. Every day Peter either walks to school. The probability that he takes a bus to school is 1/4 . If he takes a bus,the probability that he will be late is 2/3. If he walks to school, the probability that he will be late is 1/3. The probability that Peter will be ontime for at least one ou of two consecutive days is.....................
2. There are 3 women and 5 men who will split up into two 4-person teams. The number of ways in which of ways in which this can be done is .....................
3. Five card each have a single digit written on them. The digits are 9, 9, 8, 7, 6 respectively. The number of different 4- digit numbers that can be formed by placing four the cards side by side is................
4.The number of arrangements of all the eleven letters of the word
M I S S I S S I P I
In which all the four letters I are consecutive is equal to .,.....................
5. Die A has 4 red faces and 2 green faces, whereas die B has 2 red faces and 4 green faces.
A fair coin is flipped once. If it lands on a head, die A is tossed; otherwise, die B is tossed. Subsequently, the same die is tossed two more times.
a. Find the probability of obtaining a red face in firs throw!
b. Find the probability that red faces turn up in the first two throws!
c. What is the probability that no two consecutive green faces occur in the three throws?
d. In the first two throws are both red, find the probability that:
i. A red face is ontaineat the third throw.
ii. It is die A that is being tossed.
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ReplyDeleteEx1.
ReplyDeleteB : Peter takes the bus
nB : (not-B : Peter walks)
nT : (not on time) Peter is LATE !!
T : Peter is on time
so we have :
P(nT | B) = 2/3
P(nT | nB) = 1/3
P(B) = 1/4
P(nB) = 1 - P(B) = 3/4
the total proba theorem gives :
P(nT) = P(nT | B)*P(B) + P(nT | nB)P(nB)
= 2/3 * 1/4 + 1/3 * 3/4
= 5/12
as T can be envisaged in T1 and T2 independent variables for day 1 and day 2
P( Peter is late for two consecutive days) = P(nT1) * P(nT2) = (5/12)^2 = 25/144
so
P( Peter is on time for at least one out of two consecutive days) = 1 - 25/144
= 119/144
# 82.6%
Ex2.
Chosen 4 in 8 person that can be formed by no placing: C^8_4
Chosen 4 in 4 person that can be formed by no placing: C^4_4
=> C^8_4 x C^4_4 = 70
Ex3.
Chosen 4 in 5 digit number that can be formed by placing (9,9,8,7,6): A^5_4
Arrangement 2 in 4 location that can be formed by not placing (9,9): C^4_2
Chosen 2 in 3 digit number, then arrangement 2 location that can be formed by placing (8,7,6): A^3_2
=> A^5_4 – C^4_2 x A^3_2 = 60
Ex4.
(8!)/(2!x4!)