Three coins are tossed. If one of them shows tail, then find the probability that all three coins show tail.
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A bag contains `(2n+1)` coins. It is known that `n` of these coins have a head on both sides whereas the rest of the coins are fair. A coin is picked
Given , n coins have head on both sides and (n+ 1) coins are fair coins . Let `E_(1)` = Event that an unfair coin is selected `E_(2)` = Event...
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If eight coins are tossed together, then the probability of getting exactly 3 heads is
We know that, probability distribution P(X=r)=`""^(n)C_(r)(p)^( r)q^(n-r)` Here, n=8,r=3,p=`1/2` and q=`1/2` `therefore` Required probability=`""^(8)C_(3)(1/2)^(8-3)=(8!)/(5!3!)(1/2)^(8)` `=(8cdot7cdot6)/(3cdot2)cdot1/(16cdot16)=7/32`
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Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail then throw a die. Find the conditional probabili
According to the equation, sample space `S = {(H,H),(H,T),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}` In the above set, there are 8 elementary events, but all are not equally likely. However, events (H,H) and (H,T) are...
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A bag contains 3 red, 7 white, and 4 black balls. If three balls are drawn from the bag, then find the probability that all of them are of the same co
Correct Answer - `(10)/(91)` The required probability is `(.^(3)C_(3) + .^(7)C_(3) + .^(4)C_(3))/(.^(14)C_(3)) = (1 + 35 + 4)/(14 xx 13 xx 2) = (40)/(14 xx 26) = (10)/(91)`
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A fair coin is tossed repeatedly. If tail appears on first four tosses, then find the probability of head appearing on fifth toss.
Since the trials are independent, so the probability that head appears on the fifth toss does not depend upon previous results of the tosses. Hence, required probability is equal to...
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A box contains N coins, m of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is 1/2 while it is 2
Let `E_(1)` be the event that the coin drawn is fair and `E_(2)` e the event that the coin drawn is biased. Therefore, `P(E_(1))=m/NandP(E_(2))=(N-m)/(N)` A is the event that on...
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A fair coin is tossed `n` times. if the probability that head occurs 6 times is equal to the probability that head occurs 8 times, then find the value
Correct Answer - `n=14` According to question `""^(n)C_(6)((1)/(2))^(6)((1)/(2))^(n-6)=""^(n)C_(8)((1)/(2))^(8)((1)/(2))^(n-8)` `or ""^(n)C_(6)((1)/(2))^(n)=""^(n)C_(8)((1)/(2))^(n)` `or""^(n)C_(6)=""^(n)C_(8)=""^(n)C_(n-8)` `implies6=n-8orn=14`
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Let E and F be two independent events. The probability that exactly one of them occurs is `11//25` and the probability of none of them occurring is `2
Correct Answer - A::D Let `P(E)=eand P(F)=f` `P(EuuF)-P(EnnF)=11/25` `impliese+f-2ef=11/25" "(1)` `P(barEnnbarF)=2/25` `implies(1-e)(1-f)=2/25" "(2)` From (1) and (2), `ef=12/25and e+f=7/5` Solving, we get `e=4/5,f=3/5or e=3/5,f=4/5`
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A biased coin with probability `p(0 lt plt 1)` of falling tails is tossed until a tail appears for the first time. If the probability that tail comes
Correct Answer - D `(d)` `P(T)=p`, `P(H)=1-p` `:.` Required probability `=P(T or HHT or HHHHT or ….)` `=p+(1-p)^(2)p+(1-p)^(4)+….` `=(p)/(1-(1-p)^(2))` `=(p)/(2p-p^(2))=(2)/(3)` `impliesp=(1)/(2)`
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Three coins are tossed 100 times, and three heads one head occurred 14 times and head did not occur 23 times. Find the probability of getting more tha
Correct Answer - `(63)/(100)`
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