Probability Questions

Solution : Let the probability of getting a head be p and not getting a head be q. Since, head appears first time in an even throw 2 or 4 or 6. $\therefore$   $2\over 5$ = qp + $q^3$p + $q^5$p + …… $\implies$  $2\over 5$ = $qp\over {1 – q^2}$ Since q = 1- […]

Solution : Let $A_1$, $A_2$, $A_3$ be the events of match winning in first, second and third matches respectively and whose probabilities are P($A_1$) = P($A_2$) = P($A_2$) = $1\over 2$ $\therefore$  Required Probability = P($A_1$)P($A_2’$)P($A_3$) + P($A_1’$)P($A_2$)P($A_3$) = $({1\over 2})^3$ + $({1\over 2})^3$  = $1\over 8$  + $1\over 8$ = $1\over 4$ Similar Questions

Solution : Probability of getting success, p = $1\over 6$ and probability of failure, q = $5\over 6$ Now, we must get 2 sixes in seven throws, so probability is $^7C_2$$({1\over 6})^2$$({5\over 6})^5$ and the probability that 8th throw is $1\over 6$. $\therefore$   Required Probability = $^7C_2$$({1\over 6})^2$$({5\over 6})^5$$1\over 6$ = $^7C_2\times 5^5\over {6^8}$ Similar

Question : If A and B are two mutually exclusive events, then (a) P(A) < P(B’) (b) P(A) > P(B’) (c) P(A) < P(B) (d) None of these Solution : Since, A and B are two mutually exclusive events. $\therefore$    A $\cap$ B = $\phi$ $\implies$ Either A $\subseteq$ B’ or B $\subseteq$ A’

Solution : The total number of ways in which numbers can be choosed = 25*25 = 625 The number of ways in which either players can choose same numbers = 25 $\therefore$ Probability that they win a prize = $25\over 625$ = $1\over 25$ Thus, the probability that they will not win a prize in

Solution : Since the probability of solving the problem by A, B and C is $1\over 2$, $1\over 3$ and $1\over 4$ respectively. $\therefore$  Probability that problem is not solved is = P(A’)P(B’)P(C’) = ($1 – {1\over 2}$)($1 – {1\over 3}$)($1 – {1\over 4}$) = $1\over 2$$2\over 3$$3\over 4$ = $1\over 4$ = Hence, the

Solution : The probability that Mr A selected the lossing horse = $4\over 5$ $\times$ $3\over 4$ = $3\over 5$ The probability that Mr A selected the winning horse = 1 – $3\over 5$ = $2\over 5$ Similar Questions Consider rolling two dice once. Calculate the probability of rolling a sum of 5 or an

Solution : Sample space of rolling two dice = 36 Now, Event of sum of 5 = { (1,4) (2,3) (3,2) (4,1) } = 4 $\implies$ probability of getting sum of 5 is 4/36 = 1/9 Now, Event of even sum = { (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6)

Solution : Given, probabilities of speaking truth are P(A) = $4\over 5$ and P(B) = $3\over 4$ And their corresponding probabilities of not speaking truth are P(A’) = $1\over 5$  and P(B’) = $1\over 4$ The probability that they contradict each other = P(A).P(B’) + P(B).P(A’) = $4\over 5$ $\times$ $1\over 4$ + $1\over 5$

Solution : $\therefore$  Total number of cases = $3^3$ Now, favourable cases = 3(as either all has applied for house 1 or 2 or 3) $\therefore$ Required Probability = $3\over 3^3$ = $1\over 9$ Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence from match to