1pls. xplain subash,
what i did was made 1st event successful and the rest to fail...like that i repeated the pattern.....
My max sir gave some questions and i donno whether the answers r rite & method 4 some is unknown 2 me....
1. A die is thrown thrice, find the probability of getting an odd number at least once
Answer got: 5/8
2. The probabilities of happening of 3 independent events are p1, p2, p3. Find the probability that atleast one of the events will happen.
Answer got: 1-[(1-p1).p2.p3 + p1. (1-p2).p3 + p1.p2. (1-p3)]
3. A bag contains 3 red, 4 black and 2 green balls. Two balls are drawn at random from the bag. Find the probability that both balls are diff. colour.
Answer got: 13/18
4. Let A and B be two events such that P(A’ n B)=0.15 and P( A n B)=0.10. Find P(A/B).
I donno the method ( I used venn diagram and got P(B) as 0.25)
5. In 3 throws of a pair of dice, find the probability of throwing a doublet more than twice.
Method please….
6. A random variable X has the following distribution
X -2 -1 0 1 2 3
P(X) 0.1 K 0.2 2K 0.3 K
Should we include P(X) for X=-2 and -1
7. P(A’) =0.7 , P(B) =0.7 , P(B/A) =0.5, P(A/B)= ?? Answer got: 3/14 Doubt : P(A n B) > 1 (got in intermediate step)
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UP 0 DOWN 0 0 17
17 Answers
11prob should be same as
1-( prob of no event happening ie all three fail)
n u havent included cases like 2 events are successful i hope u understood ur mistake
11one more Q,
A bag contains 3 white, 5 red and 2 black balls. If 3 balls are drawn one by one with replacement, find the probability that none is white
I got it as (7/10)3
13probabilty of getting the same no. in the 2nd throw as in 1st is 1/6
isn't it??[7]
so...probability of getting a doublet is 1/6[7]
13bhiyaa...no. 5....
watz the probab of getting a single doublet???
137.
P(A)=1-0.7=0.3
P(A∩B) / 0.3= 0.5
P(A∩B)=0.15...it is <1 dude
P(A/B)=P(A∩B)/P(B)=0.15/0.7=3/14
621. A die is thrown thrice, find the probability of getting an odd number at least once
Answer got: 5/8
=1 - probability of not getting an odd number ever.
Probability of not getting an odd number on a single throw is 1/2
Probability of not getting an odd number on three throws is (1/2)3
Thus, given probability = 1 - (1/2)3
2. The probabilities of happening of 3 independent events are p1, p2, p3. Find the probability that atleast one of the events will happen.
Answer got: 1-[(1-p1).p2.p3 + p1. (1-p2).p3 + p1.p2. (1-p3)]
1- probab of no event happening.
=1 - (1-p1)(1-p2)(1-p3)
3. A bag contains 3 red, 4 black and 2 green balls. Two balls are drawn at random from the bag. Find the probability that both balls are diff. colour.
Answer got: 13/18
this is 1- probab of picking the same balls. (as by abhirup)
4. Let A and B be two events such that P(A’ n B)=0.15 and P( A n B)=0.10.
Find P(A/B).
I donno the method ( I used venn diagram and got P(B) as 0.25)
P(A/B)=P(A∩B)/P(B)
P(B)=P(A∩B)+P(A'∩B)
so answer = .1/(.1+.15) = 2/5
5. In 3 throws of a pair of dice, find the probability of throwing a doublet more than twice. Method please….
= Probab of doublet thrice.
= (1/6)3 (mistake removed)
6. A random variable X has the following distribution
X -2 -1 0 1 2 3
P(X) 0.1 K 0.2 2K 0.3 K
Should we include P(X) for X=-2 and -1
Yes you should. It is over all state space.
7. P(A’) =0.7 , P(B) =0.7 , P(B/A) =0.5, P(A/B)= ??
Answer got: 3/14
Doubt : P(A n B) > 1 (got in intermediate step)
how is P(AnB)>1?? con you show the intermediate step?
P(B/A)=P(anB)/P(a)
P(BnA)=P(B|A)/P(a) = .5 x (1-.7) = .15
P(A/A)=P(anB)/P(B) = .15/.70 = 3/14
135. prob of getting a doublet is 1/6
so prob of getting doublet thrice =1/6*1/6*1/6=1/216
131.
i m getting 7/8[2]
prob. of getting no odd no. in three throws=1/2*1/2*1/2=1/8
so ans=1-1/8=7/8
133.
P=1 - (3/9*2/8 + 4/9*3/8 + 2/9*1/8)
=13/18
ans right!!