Students and Teachers Forum

Here,         First quartile (Q1) = 30.625, so Q1 class = 30 – 40.                     Cumulative frequency table of given data is given .....

   The maximum marks obtained by below 75% students = Q3.                Cumulative frequency table of given data is given by:

We have given,                  = a and ∑fx=b,                 Number of terms (N) = ? We have,     .....

Given         ∑fx = 370 + 20a, ∑f  = 23 + a and  = 17.         a = ? We have,  Mean () = ∑fx∑f = 370+20a23+a Or, 17 = 370+20a23+a Or, 391 + 17a = 370 + 20a Or, 21 = 3a or, a = .....

Given,              = 50 and ∑ fx = 750,              Number of terms (N)= ? We have,  = ∑fxN or, 50=750N ∴N=75050=15, which is required .....

Here,   Daily wages (Rs.) No. of workers (f) Cumulative frequency (cf) 30-40 3 3 40-50 4 7 50-60 12 19 60-70 8 27 70-80 5 32   N = .....

Given,             First quartile class = 20- 30 and if (N4 - c.f) =8, We have given,  The first quartile of the given series is given by,          .....

Given,          First quartile class = 20- 30 and if (N4 - c.f) =8. The first quartile of the given series is given by,           First quartile = L + if (N4 - .....

Given,         Third quartile class = 40 – 50 and if (3N4 - c.f)=4. The third quartile of the given series is given by,                  Q3 = L + if (3N4 - .....

Given,          = 15, N = 32 and ∑fx = 350+13p,         p = ? We have,  = ΣfxN             or, 15 = 350+13p 32   or, 350+13p = 32×15   or, 13p .....

In the given curve, sum of frequencies (N) = 20. We have, Third quartile class = (3N4 )th item                                   = (34 ⨉ 20)th item   .....

We have given,             In a continuous series, the value of assumed mean is a and the sum of frequencies (f) is N &the deviation is (d),             .....

In the given curve, sum of frequencies (N) = 30. We have, median class = (N2 )th item                        = (302)th item                  .....

Given,            Mean () = 12, ∑fx = 70 + 10a and the number of frequency (N) = 5 + a,            a = ? We have,  Mean () = ∑fxN = 70+10a5+a Or, 12 = 70+10a5+a Or, .....

Given,          = 50,          ∑fx = 750,         Number of terms (N) = ? We have,  = ∑fxN or, 50=750N ∴ N=75050=15, which is required .....

In the given curve, sum of frequencies (N) = 20. We have, First quartile class =( N4)th item                                = (20 4)th item       .....

Given,          =32, N = 15 and ∑fx = k,         k = ?. We have,         = Σfx               .....

Solution: The sample space when two dice are thrown is:  

1st dice/ .....

When a coin is tossed two times, the probability of getting head in 1st toss P(H) = 12  And the probability of getting head on 2nd toss is P(H) = 12 ∴ The probability of getting both head when a coin is tossed is       .....

Here, Total no. of trails n(S) = 1200 i.    Let E1 be the event of getting faces less than 4.       Then            n(E1) = 186 + 205 + 211               .....

Let, G, Y and B denote the events of getting a ball of green, a ball of yellow and a ball of blue respectively. Then, the tree diagram is as shown in .....

Let, W and B denote the events of getting a ball of white and a ball of black respectively. Then, the tree diagram is as shown in figure. Now, from the tree diagram,     We have, probability of getting same colours= P(BB) + P(WW) = .....

Here,        The probability of solving a problem by student X, P(X) = 80 % = 0.8        The probability of solving a problem by student Y, P (Y) = 75% = 0.75. Now,         The .....

Here,          Sample space(S) = {1, 2, 3, 4, 5, 6}                              ∴n(S) = 6. Now, probability of getting 1, 3 or 5 is given .....

Let, H and T denote the events of getting a head and a tail respectively. Then, the tree diagram is as shown in figure. Then from the diagram, probability that both are tails = P(TT) = .....

Let, S = son and D = daughter. This is the case of independent event.          Probability of a son, P(S) = 12          Probability of a daughter, P(D) = 12. Tree-diagram: Sample space: .....

Let, W = white ball, B = blue ball and G = green ball Then,         n(W) = 3, n(B) = 4 and n(G) = 5 This is the case of dependent event. Tree-diagram:  From the tree diagram,  P(both are same color) = P(WW) or .....

For cards;           Total no. of cards n(S) = 52 .  Let E1 be the event of getting king, then n(E1) = 4. ∴ Probability of getting king cards P(E1) =  452             .....

No. of head = n(H) = 1 No. of tail = n(T)= 1. Tree diagram: Outcomes:            Probability: (H, H, H )             P(H, H, H)   = 1/2 × 1/2 × 1/2 = 1/8   .....

Here,         No. of red ball     = 1        No. of blue ball   = 1        No. of white ball = 1. The Probability of getting one ball of each colour is: P(R, B, W) + .....

    Let A and B be the event that the 1st and 2nd person hit the target. Now,             P(A) = 34            P(A)ˈ = 1 - 34  = 14.           .....

Let, B = black pen, Y = yellow pen and W = white pen Then, n(B) = 5, n(Y) = 3 and n(W) = 2 Probability of an ace card, P(A) = 452  Probability of a non-face card, P(N) = 4852. Tree-diagram: From the diagram, a) Both are blue = .....

Given,        The probability that A can solve a problem is 15 . We know,        Probability that A can not solve = P(A') = 1 - P(A)                   .....

Let, W = white ball, B = blue ball and G = green ball     Then,             n(W) = 1, n(B) = 1 and n(G) = 1 This is the case of dependent event. Tree-diagram: Sample space = {GB, GW, BG, BW, WG, .....

Let, B = black balls and W = white balls. Then,        n(R) = 6 and n(W) = 3. Tree-diagram: This is the case of dependent event. a) From the tree diagram, P(both white) = P(WW)               .....

Let, B = blue balls, Y = yellow balls and W = white balls. Then,          n(B) = 5, n(Y) = 3 and n(W) = 2. Probability of an ace card, P(A) =  452 Probability of a non-face card, P(N) = .....

Let A = ace cards and N = non-ace cards Then, n(A) = 4 and n(N) = 48 Tree-diagram: This is the case of dependent event. From the tree diagram, P(both aces) = P(AA) = 452 X 351                    .....

Let, G = green  and B = blue Then, n(G) = 6 and n(B) = 3 This is the case of dependent event. Probability of a green ball, P(G) = 69 Probability of a blue ball, P(B) = 39 Tree-diagram: a) Sample space: {BB, BG, GG, GB}. b) Probability .....

Solution: Let, R = red balls  and W = white balls. Then, n(R) = 3 and n(W) = 2. Tree-diagram: This is the case of independent event. Probability of a red ball, P(R) = 35 Probability of a white ball, P(W) = 25 From the tree .....

Let, G and B denote the events of giving birth to a girl and a boy respectively. Then, the tree diagram is as shown in figure. From the diagram, probability that all are girls= P(GGG) = .....

Let, B and G denote the events of selecting a boy and a girl respectively. Then, the probabilities are shown in tree diagram. Possible outcomes = { GG, GB, BG, .....

Let , B = blue  and R = red Then, n(B) = 8 and n(R) = 4 This is the case of dependent event. Probability of a blue ball, P(B) = 812 Probability of a red ball, P(R) = 412 Tree-diagram: a) Probability that both balls are blue, P(BB) = P(B) .....

Here, for a dice,           Sample space, S1 = {1, 2, 3, 4, 5, 6}. For a coin,           Sample space, S2 = {Head, tail}. Now, probability of getting 3 on the dice and head on the coin is .....

Let ,W = white  and G = green Then, n(W) = 3 and n(G) = 6 This is the case of dependent event. Probability of a white ball, P(W) = 39  Probability of a green ball, P(G) = 69 Tree-diagram: a) Probability that one is white and the .....

Let A be the event of king cards and B be the events of black cards.  Then,         Number of king cards n(A) = 4          Number of black cards n(B) = 26          Number of .....

Let A and B be the set of balls numbered with multiple of 4 and 5 respectively.  Then,          S = {1, 2, ........., 20}         A = {4, 8, 12, 16, 20}          B = {5, 10, .....

Let A and B be the set of cards divisible by 4 and 7 respectively.  Then,        S = {2, 3, 4, ..........., 25}         A = {4, 8, 12, 16, 20, 24}         B = {7, 14, .....

Given,         Number of yellow balls = 5          Number of blue balls = 3          Number of green balls = 2          Total number of balls = 5 + 3 + .....

Given,         No. of face cards = 12         No. of non face cards = 40         Total no. cards = 52.      Let E be the event of getting face cards and Eˈ be the event of .....

Let A and B are sets of prime number and cube number cards respectively.           ∴ S = {6, 7, 8, ............, 39}             A = {7, 11, 13, 17, 19, 23, 29, 31, 37}     .....