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Exercise 11.5 · Q2

Q.The probability that Mr. Q hits a target at any trial is 14\dfrac14. Suppose he tries at the target 1010 times. Find the probability that he hits the target

(i) exactly 44 times
(ii) at least one time.
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X=X= number of hits in 1010 independent trials with p=14⇒X∼B(10,14)p=\tfrac14\Rightarrow X\sim B(10,\tfrac14); part (i) is a direct pmf substitution, part (ii) uses the complement of "zero hits".

Step 1. Identify the distribution. n=10, p=14, q=34n=10,\ p=\dfrac14,\ q=\dfrac34; X∼B ⁣(10,14)X\sim B\!\left(10,\dfrac14\right).

Step 2. (i) Exactly 44 hits. P(X=4)=(104)(14)4(34)6=210×1256×7294096=210×7291048576=1530901048576≈0.1460P(X=4)=\dbinom{10}4\left(\dfrac14\right)^4\left(\dfrac34\right)^6=210\times\dfrac1{256}\times\dfrac{729}{4096}=\dfrac{210\times729}{1048576}=\dfrac{153090}{1048576}\approx0.1460.

Step 3. (ii) At least one hit — use the complement. P(X≥1)=1−P(X=0)=1−(100)(14)0(34)10=1−(34)10P(X\ge1)=1-P(X=0)=1-\dbinom{10}0\left(\dfrac14\right)^0\left(\dfrac34\right)^{10}=1-\left(\dfrac34\right)^{10}.

Step 4. Evaluate (3/4)10(3/4)^{10}. (34)10=310410=590491048576≈0.0563\left(\dfrac34\right)^{10}=\dfrac{3^{10}}{4^{10}}=\dfrac{59049}{1048576}\approx0.0563.

Step 5. Finish. P(X≥1)≈1−0.0563=0.9437P(X\ge1)\approx1-0.0563=0.9437.

✓Final answer

  1. P(X=4)≈0.1460P(X=4)\approx0.1460.
  2. P(X≥1)=1−(34)10≈0.9437P(X\ge1)=1-\left(\dfrac34\right)^{10}\approx0.9437.

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